Questions for Short Answer
1. If neutrons exert only attractive force, why don't we have a nucleus containing neutrons alone?
ANSWER: The nucleus is at the center of an atom where most of its mass is concentrated whereas the rest of its mass is associated with electrons that revolve around the nucleus. Electrons are bound to the nucleus due to the electrostatic force of attraction because the electrons have a negative electric charge and protons at the nucleus have a positive charge.
The neutrons have an attractive force due to the close-range nuclear forces which decrease very rapidly with the distance. The effect of this force is so negligible at the distance of electrons that they can not revolve around the neutrons under this force because there is no electrostatic force of attraction due to electrically neutral neutrons. So only neutrons can not make a nucleus of an atom.
2. Consider two pairs of neutrons. In each pair, the separation between the neutrons is the same. Can the force between the neutrons have different magnitudes for the two pairs?
ANSWER: If the two pairs of neutrons are far away from the influence of other particles, then the magnitude of forces between them will be the same. It may differ if each of the pairs is part of some nucleus.
3. A molecule of hydrogen contains two protons and two electrons. The nuclear force between these two protons is always neglected while discussing the behavior of a hydrogen molecule. Why?
ANSWER: A molecule of hydrogen contains two hydrogen atoms that are bound to each other due to sharing of one orbiting electron of each atom. So the distance between two protons is about the diameter of a hydrogen atom. At this distance, the close-range attractive nuclear force has no effect. Due to the presence of an electron in the orbit which has an equal and opposite electric charge, a hydrogen atom is electrically neutral and the repulsive electrostatic force between the protons is neutralized by the orbiting electrons.
So the nuclear force between the two protons in a hydrogen molecule is always neglected while discussing the behavior of a hydrogen molecule.
4. Is it easier to take out a nucleon from carbon or from iron? From iron or from lead?
ANSWER: Binding energy per nucleon is the energy required to take out a nucleon from a nucleus. We know that this energy is highest in the zone of mass number 50 to 80. It decreases before and after this zone. Iron has a mass number of 56, so it falls in the above zone. Carbon's mass number is 12 whereas lead's is 207. These two are away from the above zone.
Thus between carbon and iron, iron has more binding energy per nucleon. So it is easier to take out a nucleon from carbon than iron.
Similarly, the binding energy per nucleon for lead is less than the iron. So it is easier to take out a nucleon from lead than iron.
5. Suppose we have 12 protons and 12 neutrons. We can assemble them to form either a 24Mg nucleus or two 12C nuclei. In which of the two cases more energy will be liberated?
ANSWER: Due to the mass number, the binding energy per nucleon (say X) for Mg will be more than that for carbon (say Y). So the binding energy for Mg will be 24X. The binding energy of a carbon nucleus =12Y. For two carbon nuclei, total binding energy =24Y.
Since X > Y, →24X > 24Y. So one Mg nucleus has more binding energy than two C nuclei. Mg nucleus will release more energy.
6. What is the difference between cathode rays and beta rays? When the two are traveling in space, can you make out which is the cathode ray and which is the beta ray?
ANSWER: Both are formed with electrons but their origin is different. Cathode rays are formed in a cathode tube by accelerating the thermal electrons with a high potential difference. Beta rays are formed due to the disintegration of nuclei where neutrons convert into protons, emitting electrons.
Some electrons in a beta ray have high energy of the order of MeV that has some penetration power. Cathode rays have relatively low energy electrons that depend upon the potential difference applied across the cathode tube. They may be recognized by testing their energies in space.
7. If the nucleons of a nucleus are separated from each other, the total mass is increased. Where does the mass come from?
ANSWER: Mass is a form of energy and they are related by E =mc². When the nucleons of a nucleus are separated, energy equal to the binding energy is required to be given to the nucleons. This energy is converted into mass, which is why the total mass is increased.
8. In beta decay, an electron (or a positron) is emitted by a nucleus. Does the remaining atom get oppositely charged?
ANSWER: Suppose an electron is emitted in beta decay, the number of protons in the nucleus increases by one forming a new element. So the new atom requires one more electron in its orbit to remain electrically neutral. Since the orbiting electrons are unchanged, the remaining atom becomes an ion and is oppositely charged.
9. When a boron nucleus (105B) is bombarded by a neutron, an α-particle is emitted. Which nucleus will be formed as a result?
ANSWER: The boron nucleus has 5 protons and 5 neutrons. When it is bombarded by a neutron, it emits an α-particle. An α-particle has two protons and two neutrons. So with the emission of an α-particle, two protons are reduced in the nucleus. There remains only 3 protons. So the new atomic number of the atom is 3 now, thus forming a Li atom.
10. Does a nucleus lose mass when it suffers gamma decay?
ANSWER: No. The nucleons inside a nucleus move in discrete quantum states with different energies. In the ground state, the nucleons have minimum energy. They require very high energy, of the order of MeV's to go to higher excited states which is very difficult to provide. But in radioactive decay with α-particle or ß-particles, the newly formed daughter nucleus is generally in one of the excited states. The nucleons in the excited state eventually come to the ground state by releasing a photon (or photons) of electromagnetic radiation (similar to the process in a hydrogen atom when an electron jumps down to a lower energy state). These emitted photons by a nucleus have very high energy and are called gamma radiation and the process is gamma decay. As we have seen that gamma decay is only due to the change in energy state of the nucleus, no change of mass is involved. So the nucleus does not lose mass in a gamma decay.
11. In a typical fission reaction, the nucleus is split into two middle-weight nuclei of unequal masses. Which of the two (heavier or lighter) has greater kinetic energy? Greater linear momentum?
ANSWER: In the given situation we consider only two unequal split masses. We assume that the original nucleus is at rest i.e. with zero momentum. From the conservation principle of momentum, after fission, total momentum should also be zero. Thus both fragments will have equal linear momentum but opposite in direction.
The relation between momentum and kinetic energy is given as,
K.E. =½mv² =½(mv)²/m =p²/2m.
Since p is the same, the fragment having a lesser mass will have greater kinetic energy.
12. If three helium nuclei combine to form a carbon nucleus, energy is liberated. Why can't helium nuclei combine on their own and minimize energy?
ANSWER: A helium nucleus contains two protons and two neutrons, it is positively charged. The nucleus is bound together due to attractive close-range nuclear force among the nucleons. The range of this force is very very short and decreases rapidly with distance. The positively charged helium nuclei repel each other with the electrostatic force which is a long-range force. So this repulsion from a distance will never let the helium nuclei come so near to each other where close-range attractive nuclear force is appreciable. Thus they are never able to combine on their own to form a carbon nucleus and minimize energy.
OBJECTIVE-I
1. The mass of a neutral carbon atom in the ground state is
(a) exact 12 u
(b) Less than 12 u
(c) More than 12 u
(d) depends on the form of carbon such as graphite or charcoal.
ANSWER: (a).
EXPLANATION: Unified atomic mass unit is denoted by 'u' which is equal to 1/12th of the mass of a neutral carbon atom in its lowest energy state. So by definition, the mass of a neutral carbon atom is taken exactly as 12 u. Option (a) is correct.
2. The mass number of a nucleus is equal to
(a) The number of neutrons in the nucleus
(b) The number of protons in the nucleus
(c) The number of nucleons in the nucleus
(d) None of them.
ANSWER: (c).
EXPLANATION: The mass number A is the total number of protons Z and neutrons N (i.e. nucleons) in a nucleus. A =Z +N. Hence option (c) is correct.
3. As compared to ¹²C atom, ¹⁴C atom has
(a) two extra protons
(b) two extra protons but no extra electron
(c) two extra neutrons and no extra electron
(d) two extra neutrons and two extra electrons.
ANSWER: (c).
EXPLANATION: ¹⁴C is an isotope of carbon. Its nucleus has the same number of protons as ¹²C (otherwise if the number of protons changes, the element will change and it will not remain a C atom). But it has two extra neutrons that's why its mass number is 12+2 =14. In a neutral atom, the number of electrons and the number of protons are the same. Hence ¹⁴C has the same number of electrons as ¹²C. Option (c) is correct.
4. The mass number of a nucleus is
(a) always less than its atomic number
(b) always more than its atomic number
(c) equal to its atomic number
(d) sometimes more than and sometimes equal to its atomic number.
ANSWER: (d).
EXPLANATION: The mass number =Atomic number Z (number of protons) + N (number of neutrons.
So the mass number can not be less than its atomic number but may be more than or equal to it. It is because of the fact that in a nucleus Z cannot be zero. But N may be zero as is the case with hydrogen atoms where Z=1, N=0, and A =1.
So the mass number is sometimes more than and sometimes equal to its atomic number. Option (d) is correct.
5. The graph of ln (R/Rₒ) versus ln A (R = radius of a nucleus and A = its mass number) is
(a) a straight line
(b) a parabola
(c) an ellipse
(d) None of them.
ANSWER: (a).
EXPLANATION: The average radius R of a nucleus is given as
R =RₒA1/3
Where Rₒ =1.1 fm.
→R/Rₒ =A1/3
→ln (R/Rₒ) =(1/3) ln A
If y =ln (R/Rₒ) and x =ln A
then, y =(1/3)x
It is in the form of y =mx, which is an equation of a straight line passing through the origin and having slope =1/3.
Option (a) is correct.
6. Let Fₚₚ, Fₚₙ, and Fₙₙ denote the magnitude of the nuclear force by a proton on a proton, by a proton on a neutron, and by a neutron on a neutron respectively. When the separation is 1 fm,
(a) Fₚₚ > Fₚₙ = Fₙₙ
(b) Fₚₚ = Fₚₙ = Fₙₙ
(c) Fₚₚ > Fₚₙ > Fₙₙ
(d) Fₚₚ < Fₚₙ = Fₙₙ.
ANSWER: (b).
EXPLANATION: At the given separation of 1 fm, the close-range attractive nuclear force is predominant which is the same for each nucleon pair and independent of charge. Hence option (b) is correct.
7. Let Fₚₚ, Fₚₙ, and Fₙₙ denote the magnitude of the net force by a proton on a proton, by a proton on a neutron, and by a neutron on a neutron respectively. Neglect gravitational force. When the separation is 1 fm,
(a) Fₚₚ > Fₚₙ = Fₙₙ
(b) Fₚₚ = Fₚₙ = Fₙₙ
(c) Fₚₚ > Fₚₙ > Fₙₙ
(d) Fₚₚ < Fₚₙ = Fₙₙ.
ANSWER: (d).
EXPLANATION: The net force between a nucleon pair at the given distance will be the sum of nuclear force and coulomb force. The coulomb force here will be only between a proton-proton pair because only a proton has a positive charge while the neutron is neutral. So the net force between the proton-neutron pair and the neutron-neutron pair will be equal. The coulomb force between a proton-proton pair will be repulsive while the nuclear force between them is attractive. So the net force between a proton-proton pair will be less than the proton-neutron pair or neutron-neutron pair.
Option (d) is correct.
8. Two protons are kept at a separation of 10 nm. Let Fₙ and Fₑ be the nuclear force and the electromagnetic force between them.
(a) Fₑ = Fₙ,
(b) Fₑ >> Fₙ,
(c) Fₑ << Fₙ,
(d) Fₑ and Fₙ differ only slightly.
ANSWER: (b).
EXPLANATION: Nuclear force is only appreciable when separation is of the order of fm. The nm is a very large unit compared to the fm. The nuclear force is a close-range force that reduces drastically with an increase in separation and becomes negligible when the distance is increased up to 10 fm. So at a separation of 10 nm, Fₙ will be nearly zero while Fₑ will have appreciable value. Thus Fₑ >> Fₙ.
Option (b) is correct.
9. As the mass number A increases, the binding energy per nucleon in the nucleus
(a) increases
(b) decreases
(c) remains the same
(d) varies in a way that depends on the actual value of A.
ANSWER: (d).
EXPLANATION: The binding energy per nucleon in the nucleus depends upon the number of surrounding nucleons around an inside nucleon and the surface area of a nucleus. In lighter nuclei surrounding nucleons inside a nucleus increase with A, so initially the B.E. per nucleon increases with A up to the point when the surrounding nucleons around a nucleon inside are full. Thereafter with a continuous increase in surface area, the B.E. per nucleon starts to decrease slowly. This behavior cannot be explained with options (a), (b), or (c). So option (d) is correct.
10. Which of the following is a wrong description of the binding energy of a nucleus?
(a) It is the energy required to break a nucleus into its constituent nucleons.
(b) It is the energy made available when free nucleons combine to form a nucleus.
(c) It is the sum of the rest mass energies of its nucleons minus the rest mass energy of the nucleus.
(d) It is the sum of the kinetic energy of all the nucleons in the nucleus.
ANSWER: (d).
EXPLANATION: Options (a), (b), and (c) give the correct description of binding energy. The energies described in these options are exactly the same. The kinetic energies of nucleons have nothing to do with the binding energy of a nucleus. So option (d) is the answer.
11. In one average life,
(a) Half the active nuclei decay
(b) Less than half the active nuclei decay
(c) More than half the active nuclei decay
(d) All the nuclei decay.
ANSWER: (c).
EXPLANATION: The average life of a nuclei is =Half-life/0.693
=1.44*Half-life.
So the average life of a nuclei is more than its half-life. Option (c) is correct.
12. In radioactive decay, neither the atomic number nor the mass number changes. Which of the following particles is emitted in the decay?
(a) proton
(b) neutron
(c) electron
(d) photon.
ANSWER: (d).
EXPLANATION: If a proton is emitted in a radioactive decay, atomic number as well as mass number changes. If a neutron is emitted, the atomic number remains unchanged but the mass number changes. If an electron is emitted, the mass number remains unchanged but the atomic number changes.
With the given condition in the problem, options (a), (b), and (c) are not correct. If a photon is emitted neither atomic number nor mass number changes as is the case with gamma radiation. Hence option (d) is correct.
13. During negative beta decay,
(a) An atomic electron is ejected
(b) An electron that is already present within the nucleus is ejected
(c) A neutron in the nucleus decays emitting an electron
(d) a proton in the nucleus decays emitting an electron.
ANSWER: (c).
EXPLANATION: During a negative beta decay, a neutron converts into a proton and an electron. This electron is emitted while the proton remains in the nucleus. It increases the atomic number of the nucleus but the mass number remains the same. Option (c) is correct.
14. A freshly prepared radioactive source of half-life 2 h emits radiation of intensity that is 64 times the permissible safe level. The minimum time after which it would be possible to work safely with this source is
(a) 6 h
(b) 12 h
(c) 24 h
() 128 h.
ANSWER: (b).
EXPLANATION: For safe working with the source, the radiation activity should be 1/64 of the initial activity. So,
A/Aₒ = 1/2t/t½
→1/2t/t½ =1/64 =1/2⁶
→t/t½ =6
→t =6*t½ =6*2 h =12 h.
Option (b) is correct.
15. The decay constant of a radioactive sample is λ. The half-life and the average life of the sample are respectively
(a) 1/λ and (ln 2/λ)
(b) (ln 2/λ) and 1/λ
(c) λ(ln 2) and 1/λ
(d) λ/(ln 2) and 1/λ.
ANSWER: (b).
EXPLANATION: The half-life is given as,
t½ =(ln 2)/λ
While the average life is given as
tₐᵥ =1/λ
Option (b) is correct.
16. An α-particle is bombarded on ¹⁴N. As a result, a ¹⁷O nucleus is formed and a particle is emitted. This particle is a
(a) neutron
(b) proton
(c) electron
(d) positron.
ANSWER: (b).
EXPLANATION: An α-particle has two protons and two neutrons. When it is absorbed by a ¹⁴N nucleus which has seven protons and seven neutrons, the new temporary nucleus has nine protons and nine neutrons and the mass number becomes 18. But the new nucleus formed is given that of oxygen which has 8 protons only. So clearly a proton is emitted and the mass number also becomes 17.
Option (b) is correct.
17. Ten grams of ⁵⁷Co kept in an open container beta-decays with a half-life of 270 days. The weight of the material inside the container after 540 days will be very nearly
(a) 10 g
(b) 5 g
(c) 2.5 g
(d)1.25 g.
ANSWER: (a).
EXPLANATION: In beta decay, an electron from the nucleus is emitted. The nucleus is composed of protons and neutrons, each of them is about 1840 times more in weight than an electron. A ⁵⁷Co nucleus has a total of 57 protons and neutrons. So in comparison to a Co nucleus, the weight of an electron is negligible. So after 540 days of beta decay, the residual weight of the Co in the container will be almost the same i.e. 10 g.
Option (a) is correct.
18. Free ²³⁸U nuclei kept in a train emit alpha particles. When the train is stationary and a uranium nucleus decays, a passenger measures that the separation between the alpha particle and the recoiling nucleus becomes x in time t after the decay. If a decay takes place when the train is moving at a uniform speed v, the distance between the alpha particle and the recoiling nucleus at a time t after the decay, as measured by the passenger will be
(a) x +vt
(b) x -vt
(c) x
(d) depends on the direction of the train.
ANSWER: (c).
EXPLANATION: Since the frame of reference is moving with uniform velocity, objects will appear to an observer moving with a velocity that is equal to the object's velocity minus the frame's velocity.
Suppose the velocity of the alpha particle =u and that of the recoiling nucleus = -u' when the train is at rest. So the relative velocity of the alpha particle relative to the recoiling nucleus is u -(-u') =u +u'. It is the velocity by which both particles will appear to separate from each other to the observer.
Now consider the train moving with a uniform velocity v. The observer will see the velocity of the alpha particle =u -v and the velocity of the recoiling nucleus =-u' -v. So both particles will appear to the observer to separate from each other with a velocity equal to
=(u -v) -(-u' -v)
=u +u'
It is the same when the train is at rest. Hence in time t after the decay, the passenger will measure the same distance x when the train is moving with a uniform velocity.
19. During a nuclear fission reaction,
(a) A heavy nucleus breaks into two fragments by itself
(b) A light nucleus bombarded by thermal neutrons breaks up
(c) A heavy nucleus bombarded by thermal neutrons breaks up
(d) Two light nuclei combine to give a heavier nucleus and possibly other products.
ANSWER: (c)
EXPLANATION: When a heavy nucleus absorbs a bombarded thermal neutron and becomes unstable, it breaks up. This nuclear reaction is called a fission reaction.
Option (c) is correct.
OBJECTIVE-II
1. As the mass number A increases, which of the following quantities related to a nucleus do not change?
(a) mass
(b) volume
(c) density
(d) binding energy.
ANSWER: (c).
EXPLANATION: With the increase in mass number A, the number of nucleons increases. Thus mass and volume both increase. Also, the binding energy of the nucleus increases because, the more the number of nucleons, the more energy is required for disintegrating the nucleus in its constituents. Hence options (a), (b), and (d) are incorrect.
Let us see what happens with density. Suppose the average mass of a nucleon =m, then the mass of a nucleus =mA.
The radius of a nucleus, R =RₒA1/3
where Rₒ =1.1 fm.
The volume of a nucleus,
V =4πR³/3 =4πRₒ³A/3
Hence the density of a nucleus,
D =mass/volume
=3mA/4πRₒ³A
=3m/4πRₒ³.
This expression for the density is a constant.
Hence option (c) is correct.
2. The heavier nuclei tend to have a larger N/Z ratio because
(a) A neutron is heavier than a proton
(b) A neutron is an unstable particle
(c) A neutron does not exert electric repulsion
(d) Coulomb forces have a larger range compared to the nuclear forces.
ANSWER: (c), (d).
EXPLANATION: In heavier nuclei, the number of protons is more. They exert repulsive Coulomb force on each other which is a long-range force. To stabilize a heavier nucleus, neutrons play a vital role as they have no electrostatic charge, they do not repel other nucleons and they only exert attractive close-range nuclear force. It resists the stronger Coulomb force in heavier nuclei and helps the nucleus remain together, so the N/Z ratio is higher.
A neutron is marginally heavier than a proton but it has nothing to do with the N/Z ratio. A neutron is not an unstable particle. Options (a) and (b) are incorrect.
Options (c) and (d) are correct.
3. A free neutron decays to a proton but a free proton does not decay to a neutron. This is because
(a) neutron is a composite particle made of a proton and an electron whereas a proton is a fundamental particle.
(b) neutron is an uncharged particle whereas a proton is a charged particle
(c) neutron has a larger rest mass than the proton
(d) Weak forces can operate in a neutron but not in a proton.
ANSWER: (c).
EXPLANATION: A free proton can not decay into a neutron because a neutron has a larger rest mass than the proton. If this process is possible then the Q-value will be negative which is not acceptable and possible. Q-value is the difference in mass of the sum of the rest masses of constituents and the rest mass of the final product.
Options (a) and (d) are incorrect, while option (b) is not a reason that a free proton does not decay to a neutron. Only option (c) is correct.
4. Consider a sample of a pure beta-active material.
(a) All the beta particles emitted have the same energy
(b) The beta particles originally exist inside the nucleus and are ejected at the time of decay.
(c) The antineutrino emitted in a beta decay has zero mass and hence zero momentum.
(d) The active nucleus changes to one of its isobars after the beta decay.
ANSWER: (d).
EXPLANATION: All beta particles do not have the same kinetic energy because this energy is shared by the beta particle and the antineutrino/neutrino emitted during the decay. Depending upon the proportion of sharing, the kinetic energy of beta particles varies. Option (a) is wrong.
Beta particles do not originally exist inside a nucleus. They are formed and emitted when inside an active nucleus either a neutron decays into a proton and electron or a proton decays into a neutron and a positron. Thus the resulting nucleus is an isobar of the previous one. Option (b) is wrong but option (d) is correct. Antineutrino emitted in a beta decay have zero mass but possess momentum. Option (c) is incorrect.
5. In which of the following decays the element does not change?
(a) α-decay
(b) ß⁺-decay
(c) ß⁻-decay
(d) 𝛾-decay
ANSWER: (d).
EXPLANATION: Only in a gamma decay the element does not change because the emitted gamma rays are not charged particles but electromagnetic radiation which is a result of the newly formed nucleus coming down to the ground state of energy.
Other given particles are charged particles and their emission changes the atomic number of the nucleus. Since the atomic number decides the element, with the emission of these charged particles the element changes.
Option (d) is correct.
6. In which of the following decays does the atomic number decrease?
(a) α-decay
(b) ß⁺-decay
(c) ß⁻-decay
(d) 𝛾-decay.
ANSWER: (a), (b).
EXPLANATION: A decrease in atomic number means a decrease in the number of protons in a nucleus. In an α-decay two protons and two neutrons are emitted from a nucleus. So the atomic number decreases by two.
In a beta-plus decay, a proton is converted into a neutron and a positron. The resulting neutron remains in the nucleus while the positron is emitted. Thus the nucleus now has one proton less resulting in a decrease in atomic number.
In a beta-minus decay, a neutron is converted into a proton and an electron. The electron is emitted as a beta-minus ray. Due to the resulting proton, the number of protons in the nucleus increases i.e. the atomic number increases.
Since gamma rays are electromagnetic radiations not charged particles, the atomic number of the element remains the same.
Clearly, options (a) and (b) are correct only.
7. Magnetic field does not cause deflection in
(a) α-rays
(b) beta-plus rays
(c) beta-minus rays
(d) gamma rays.
ANSWER: (d).
EXPLANATION: A magnetic field causes a deflection in a moving charged particle. The α-rays, beta-plus rays, and beta-minus rays are charged particles, hence they get deflected by a magnetic field.
Gamma rays are electromagnetic radiations not charged particles, so they are not deflected by a magnetic field.
Option (d) is correct.
8. Which of the following are electromagnetic waves?
(a) α-rays
(b) beta-plus rays
(c) beta-minus rays
(d) gamma rays.
ANSWER: (d).
EXPLANATION: α-rays are helium nuclei, beta-plus rays are positrons (a particle similar to an electron but with opposite charge) and beta-minus rays are electrons. All these three are particles with a certain mass. Only gamma rays are electromagnetic waves.
Option (d) is correct.
9. Two lithium nuclei in a lithium vapor at room temperature do not combine to form a carbon nucleus because
(a) A lithium nucleus is more tightly bound than a carbon nucleus
(b) A carbon nucleus is an unstable particle
(c) It is not energetically favorable
(d) Coulomb repulsion does not allow the nuclei to come very close.
ANSWER: (d).
EXPLANATION: As we know most of an atom is hollow with a heavy nucleus with a positive charge at the center. So these positively charged nuclei repel each other with Coulomb force which is a long-range force. Nuclei can combine together only if they are close enough (separations of the order of a few fm) so that close-range attractive nuclear forces are effective. As in the given case, the lithium vapor is at room temperature. If two randomly moving lithium atoms hit each other they will get orbiting electrons in one atom repelling other's with Coulomb force because they do not form a covalent bond. Even if the nuclei may overcome the orbiting electrons, both nuclei will repel each other with Coulomb force. So they can not come close enough to get the attractive nuclear force significant in magnitude.
The first three options are incorrect, option (d) is correct.
10. For nuclei with A > 100,
(a) The binding energy of the nucleus decreases on average as A increases
(b) The binding energy per nucleon decreases on average as A increases
(c) If the nucleus breaks into two roughly equal parts, energy is released
(d) If two nuclei fuse to form a bigger nucleus, energy is released.
ANSWER: (b), (c).
EXPLANATION: With increasing A, the number of nucleons increases and hence the binding energy of the nucleus increases. Option (a) is incorrect.
For nuclei A > 100, binding energy per nucleon decreases on average as A increases. It is due to the fact that inner nucleons are surrounded fully and their interactions with other nucleons reach nearly maximum. But with an increase in A, the surface area of the nucleus also increases. Surface nucleons are not fully surrounded and hence their interactions with other nucleons are not saturated. Due to this, the binding energy per nucleon decreases on average as A increases. Option (b) is correct.
When a nucleus with A > 100 breaks into two roughly equal parts, the binding energy per nucleon of these parts will be more than the parent nucleus. So if we mathematically add the binding energy of these two daughter nuclei, the sum will be more than the binding energy of the parent nucleus. Thus in this process, energy is released. Option (c) is correct.
If two nuclei with A > 100 fuse together, the resulting nucleus will have binding energy per nucleon less than the parent nuclei. Thus the binding energy of the resulting nucleus will be less than the sum of the binding energy of two parent nuclei. So in this process, energy will not be released but it will require energy. Option (d) is incorrect.
EXERCISES
1. Assume that the mass of a nucleus is approximately given by M =Amₚ where A is the mass number. Estimate the density of the matter in kg/m³ inside a nucleus. What is the specific gravity of nuclear matter?
ANSWER: mₚ =1.007276 u
Given the mass of the nucleus,
M =Amₚ
The radius of a nucleus is
R = Rₒ∛A, where Rₒ =1.1 fm.
The volume of the nucleus,
V =(4/3)πR³
=(4/3)πRₒ³A
Hence the density of the nucleus
D =M/V
=3Amₚ/4πRₒ³A
=3mₚ/4πRₒ³
=3*1.007276*1.66x10⁻²⁷/4π*(1.1x10⁻¹⁵)³ kg/m³
=3.0x10¹⁷ kg/m³.
(Because of 1 u ≈1.66x10⁻²⁷ kg. You can calculate it from the given data where 1 u =931 MeV/c²)
1 u =931*10⁶*1.602x10⁻¹⁹/(3x10⁸)² kg
=1.66x10⁻²⁷ kg.
The specific gravity of the nuclear matter,
=Density of the matter/Density of water
=3.0x10¹⁷ kg/m³/1x10³ kg/m³
=3.0x10¹⁴.
2. A neutron star has a density equal to that of nuclear matter. Assuming the star to be spherical, find the radius of a neutron star whose mass is 4.0x10³⁰ kg (twice the mass of the sun).
ANSWER: The volume of the neutron star,
V = Mass/Density
Given, M =4.0x10³⁰ kg,
Density =density of nuclear matter
=3.0x10¹⁷ kg/m³
(Calculated in the previous problem).
Hence the volume,
V =4.0x10³⁰/3.0x10¹⁷ m³
=(4/3)x10¹³ m³.
But volume of a sphere =(4/3)πR³
If we take R = radius of the neutron star then,
(4/3)πR³ =(4/3)x10¹³
→πR³ =1x10¹³
→R =∛(1x10¹³/π)
=14710 m
≈15 km.
3. Calculate the mass of an α-particle. Its binding energy is 28.2 MeV.
ANSWER: An α-particle is made of two protons and two neutrons. Hence,
2mₚ +2mₙ =mₐ +binding energy
where mₐ =mass of the α-particle.
Given binding energy =28.2 MeV
In terms of mass B.E. =28.2 Mev/c²
Hence,
mₐ =2*1.007276 u+2*1.008665 u -28.2 Mev/c²
=4.031882 u -28.2 MeV/c²
=4.031882 u -(28.2/931) u
=4.0016 u.
4. How much energy is released in the following reaction?
⁷Li + p = α + α
The atomic mass of ⁷Li =7.0160 u and that of ⁴He =4.0026 u.
ANSWER: The mass of a proton =1.007276 u.
The sum of the rest mass of Li and p on the left side,
LHS =(7.0160 +1.007276) u
=8.023276 u.
Given that the mass of an α-particle
=4.0026
So the sum of the rest mass of two α-particles on the RHS is,
=2*4.0026 u
=8.0052 u.
Since the final product on the RHS has a lesser mass than the LHS, this difference in mass is released as energy.
So the mass equivalent of energy released is
=8.023276 -8.0052 u
=0.018076 u
=0.018076*931 MeV/c²
=16.83 MeV/c²
The energy equivalent of this released energy,
=16.83 MeV.
5. Find the binding energy per nucleon of ¹⁹⁷₇₉Au if its atomic mass is 196.96 u.
ANSWER: The number of protons =79,
The number of neutrons =197 -79 =118.
The sum of the rest mass of the individual protons and neutrons,
M =79x1.007276+118x1.008665 u
=198.597274 u
But the rest mass of the Au atom is =196.96 u.
Binding energy =difference in mass in energy units
=198.597274 -196.96 u
=1.637274 u
=1.637274*931 MeV
=1524.3 MeV.
Total number of nucleons =197,
Hence the binding energy per nucleon,
=1524.3/197 = 7.74 MeV.
6. (a) Calculate the energy released if ²³⁸U emits an α-particle.
(b) Calculate the energy to be supplied to ²³⁸U if two protons and two neutrons are to be emitted one by one.
The atomic mass of ²³⁸U, ²³⁴Th, and ⁴He are 238.0508 u, 234.04363 u, and 4.00260 u respectively.
ANSWER: (a) Energy released in mass equivalent,
=mass of ²³⁸U -(mass of ²³⁴Th +mass of ⁴He),
=238.0508 -(234.04363 +4.0026) u.
=238.0508 -238.04623 u
=0.00457 u
In terms of energy, the released energy,
=0.00457*931 MeV
=4.255 MeV.
(b) Rest mass of two protons and two neutrons separately,
=2*1.007276 +2*1.008665 u
=4.031882 u
So the sum of the masses of the products,
=234.04363 +4.031882 u
=238.075512 u
Since the mass of the products is more than the mass of the parent nucleus 238.0508 u, the difference in mass has to be supplied in the form of energy for this process to happen. Required energy E =238.075512 -238.0508 u
→E =0.024712 u
=0.024712*931 MeV (in energy units)
=23.00 MeV.
7. Find the energy liberated in the reaction
²²³Ra → ²⁰⁹Pb + ¹⁴C
The atomic masses needed are as follows:
²²³Ra →223.018 u
²⁰⁹Pb →208.981 u
¹⁴C → 14.003 u.
ANSWER: Atomic mass of Ra =223.018 u,
The sum of the atomic masses of products Pb and C,
=208.981 +14.003 u
=222.984 u.
Reduction in mass =223.018 -222.984 u
=0.034 u.
This much mass is converted into energy that is liberated. The energy equivalent of this mass,
=0.034*931 MeV
=31.65 MeV.
8. Show that the minimum energy needed to separate a proton from a nucleus with Z protons and N neutrons is
ΔE =(MZ-1,N +MH -MZ,N)c²
where MZ,N =mass of an atom with Z protons and N neutrons in the nucleus and MH =mass of a hydrogen atom. This energy is known as proton-separation energy.
ANSWER: The energy supplied to separate a proton from a nucleus appears as an increased mass of the products. With the separation of a proton from a nucleus with Z protons and N neutrons, the daughter nucleus has now Z-1 protons and N neutrons. The proton's mass is approximately equal to the mass of a hydrogen atom. Hence the energy supplied,
=Increase in mass of the products
=(Massof daughter nucleus+mass of the proton) -mass of the parent nucleus
=(MZ-1,N +MH) -MZ,N
=MZ-1,N +MH -MZ,N
The energy equivalent of this mass is given from E=mc², hence the energy needed to separate a proton from a nucleus,
ΔE =(MZ-1,N +MH -MZ,N)c².
9. Calculate the minimum energy needed to separate a neutron from a nucleus with Z protons and N neutrons in terms of masses MZ,N, MZ,N-1, and the mass of the neutron.
ANSWER: Increase in the sum of the mass of the products,
ΔM=(MZ,N-1 +mN)- MZ,N
This increase in mass is due to the conversion of the supplied energy to separate a neutron.
The increase in mass ΔM is written as energy according to the equation ΔE =ΔM*c²
Hence the required energy,
ΔE =ΔMc²
=(MZ,N-1 +mN -MZ,N)c².
10. ³²P beta decays to ³²S. Find the sum of the energy of the antineutrino and the kinetic energy of the ß-particle. Neglect the recoil of the daughter nucleus. The atomic mass of ³²P =31.974 u and that of ³²S =31.972 u.
ANSWER: As we see, the mass of the atom decreases with beta decay. We take the mass of a beta particle as negligible in comparison to a proton or neutron. So this decrease in mass will appear as kinetic energy of the beta particle and antineutrino.
ΔE =ΔM*c²
=[m(³²P) -m(³²S)]*c²
=[31.974 u -31.972 u]*c²
=[0.002 u]*c²
=[0.002 u]*c²*931 MeV/c²
=1.86 MeV.
11. A free neutron beta decays to a proton with a half-life of 14 minutes. (a) What is the decay constant? (b) Find the energy liberated in the process.
ANSWER: Given that, t½ =14 minutes
=14*60 s =840 s.
(a) The decay constant 𝜆 =0.693/t½
→𝜆 =0.693/840 s⁻¹
=8.25x10⁻⁴ s⁻¹.
(b) In a beta decay of a free neutron, a beta particle which is an electron is emitted and the neutron converts to a proton. The energy liberated in this process will be the energy equivalent of the mass that disappeared in the process of decay.
ΔE =Δm*c²
=[mₙ -(mₑ +mₚ)]c²
=[1.008665-(0.0005486+1.007276)]*c²
=(0.0008404 u)*c²
=0.0008404*931 MeV
=0.782 MeV
=782 keV.
12. Complete the following decay schemes.
(a) ²²⁶₈₈Ra → α⁺
(b) ¹⁹₈O → ¹⁹₉F⁺
(c) ²⁵₁₃Al → ²⁵₁₂Mg⁺.
ANSWER: (a) With an alpha decay, The Ra atom will lose two protons and two neutrons in its nucleus.
The number of protons now =88-2 =86
Total number of nucleons =226-4 =222
Since the number of protons in a nucleus is the atomic number that is related to a particular element, the new atom with 86 protons in its nucleus is Rn.
So the decay scheme is as follows:
²²⁶₈₈Ra
↓→α⁺
²²²₈₆Rn
(b) When an oxygen atom with 8 protons and 11 neutrons decays to an F atom with 9 protons and 10 neutrons, it is clear that in the nucleus a neutron has decayed into a proton and an electron. While this proton remains in the nucleus the electron is emitted as a beta-minus particle with an antineutrino. The decay scheme is as follows:
¹⁹₈O
↓→ e⁻ +𝜈̅
¹⁹₉F⁺
(c) Here an Al nucleus with 13 protons and 12 neutrons decays to an Mg⁺ nucleus having 12 protons and 13 neutrons. Clearly, the number of nucleons remains the same but a proton in the Al nucleus breaks into a neutron and a positron (e⁺). Also in this process, a neutrino is released that is emitted with the positron.
The decay scheme is as follows:
²⁵₁₂Al
↓→e⁺ 𝜈
²⁵₁₃Mg⁺
13. In the decay ⁶⁴Cu →⁶⁴Ni +e⁺ +𝜈, the maximum kinetic energy carried by the positron is found to be 0.650 MeV.
(a) What is the energy of the neutrino which was emitted together with a positron of kinetic energy 0.150 MeV?
(b) What is the momentum of this neutrino in kg-m/s?
Use the formula applicable to a photon.
ANSWER: (a) In beta decay, the released kinetic energy is shared by the beta particle and the neutrino/antineutrino. Here a positron has a maximum kinetic energy of 0.650 MeV, hence in other decays this much kinetic energy will be shared by the positron and neutrino. If the positron has a kinetic energy of 0.150 MeV, the neutrino will have a kinetic energy,
=0.650 -0.150 MeV =0.500 MeV.
(b) Momentum of this neutrino,
p =E/c
=0.500*10⁶*1.602x10⁻¹⁹/3x10⁸ kg-m/s
=2.67x10⁻²² kg-m/s.
14. Potassium-40 can decay in three modes. It can decay by ß⁻-emission, ß⁺-emission, or electron capture. (a) Write the equations showing the end products. (b) Find the Q-value in each of the three cases. Atomic masses of ⁴⁰₁₈Ar, ⁴⁰₁₉K, and ⁴⁰₂₀Ca are 39.9624 u, 39.9640 u, and 39.9626 u respectively.
ANSWER: (a) Potassium has atomic number =19, hence there will be 19 protons and 21 electrons in the nucleus of Potassium-40.
For the decay by ß⁻-emission
In this decay, a neutron in the nucleus of the potassium converts into a proton and an electron. The proton remains in the nucleus while the electron emits as ß⁻-radiation with an antineutrino. The number of nucleons in the nucleus remains the same but now there are 20 protons and 20 neutrons. Its new atomic number is 20, so it is now a calcium atom. The equation for this beta decay may be written as follows:
⁴⁰₁₉K →⁴⁰₂₀Ca +e⁻ +𝜈̅
For the decay by ß⁺-emission
In this decay, a proton converts into a neutron and a positron. The positron (e⁺) gets emitted as ß⁺-radiation with a neutrino. The nucleus has now 18 protons and 22 neutrons, so its atomic number is 18, and its mass number remains 40. It is now an Ar atom. The equation for this ß⁺-decay may be written as follows:
⁴⁰₁₉K →⁴⁰₁₈Ar +e⁺ +𝜈
For the decay by electron capture:
In this mode of decay, an orbiting electron of the atom is captured by a proton of the nucleus to convert into a neutron. A neutrino is produced in this process and emitted. Again the daughter nucleus has a total of 18 protons and 22 neutrons. Its atomic number is 18 and its mass number is 40. So it is now an Ar atom. The equation may be written as follows:
⁴⁰₁₉K +e⁻ →⁴⁰₁₈Ar +𝜈
(b) Q-values
Q-value for the beta-minus decay:
Q =Uᵢ -Uⱼ
=[m(⁴⁰₁₉K) -m(⁴⁰₂₀Ca)]c²
=[(39.964 -39.9626) u]c²*931MeV/c²
=1.3034 MeV.
Q-value for the ₑ⁺ (beta plus) decay:
Q =Uᵢ -Uⱼ
=[m(⁴⁰₁₉K) -m(⁴⁰₁₈Ar) -2mₑ]c²
=[(39.964-39.9624)c²*931 MeV/c²-2*511 keV]
=[1.4896 MeV -1022 keV
=1489.6 -1022 keV
=467.6 keV
=0.4676 MeV.
Q-value for the electron capture process:
Q =Uᵢ -Uⱼ
=[m(⁴⁰₁₉K) -m(⁴⁰₁₈Ar) ]c²*931 MeV/c²
=(39.964 -39.9624)*931 MeV
=1.4896 MeV
≈1.490 MeV.
15. Lithium (Z =3) has two stable isotopes ⁶Li and ⁷Li. When neutrons are bombarded on the Li sample, electrons and α-particles are ejected. Write down the nuclear processes taking place.
ANSWER: When a neutron is absorbed by the nucleus of a ⁶Li atom, it changes to another stable isotope ⁷Li. So the nuclear process is:
⁶₃Li +n → ⁷₃Li
Further absorption of neutron by the ⁷₃Li converts it into ⁸₃Li. So,
⁷₃Li +n → ⁸₃Li
Since the neutron in the new nucleus has a greater proportion, this atom becomes unstable. To get stability, one of the neutrons in the nucleus converts to a proton and an electron. The converted proton remains in the nucleus while the electron is emitted as ß-radiation along with an antineutrino. The atomic number Z of the daughter nucleus is 4, hence it is a Be nucleus now. The process can be written as follows:
⁸₃Li →⁸₄Be +e⁻ +𝜈̅
To achieve further stability, the Be nucleus decays into two alpha particles as follows:
⁸₄Be →⁴₂He +⁴₂He.
16. The masses of ¹¹C and ¹¹B are respectively 11.0114 u and 11.0093 u. Find the maximum energy a positron can have in the ß⁺-decay of ¹¹C to ¹¹B.
ANSWER: The energy released in this process is the Q-value of the process.
Q-value =[m(¹¹C) -m(¹¹B) -2mₑ]c²
=[(11.0114 -11.0093 -2*0.0005486)u]c²*931 Mev/c²
=(0.0021 -0.0010972)*931 MeV
=0.0010028*931 MeV
=0.9336 MeV
=933.6 keV.
This released energy is shared by the emitted positron and the neutrino. The maximum energy a positron will have when the neutrino shares zero energy. Then whole of the energy will be taken by the positron. Hence the maximum energy a positron can have =933.6 keV.
17. ²²⁸Th emits an alpha particle to reduce ²²⁴Ra. Calculate the kinetic energy of the alpha particle emitted in the following decay:
²²⁸Th → ²²⁴Ra* + α
²²⁴Ra* → ²²⁴Ra + 𝜸 (217 keV).
The atomic mass of ²²⁸Th is 228.028726 u, that of ²²⁴Ra is 224.020196 and that of ⁴He is 4.00260 u.
ANSWER: The reduction in the rest mass of the products appears as the kinetic energy of the alpha particle.
ΔM =m(²²⁸Th) -{m(²²⁴Ra*) +m(⁴He)}
=228.028726 u-(224.020196 u+217 keV+4.00260) u
=0.00593 u -217 keV
The energy equivalent of 0.00593 u is,
=0.00593c²*931 MeV/c²
=5.52083 Mev
Hence the kinetic energy of the alpha particle,
ΔE =ΔM (interms of energy)
=5.52083 MeV -217 keV
=5.52083 -0.217 MeV
=5.304 MeV.
18. Calculate the maximum kinetic energy of the beta particle emitted in the following decay scheme:
¹²N → ¹²C* +e⁺ +𝜈
¹²C* → ¹²C +𝛾 (4.43 MeV)
The atomic mass of ¹²N is 12.018613 u.
ANSWER: m(¹²C*) =m(¹²C) +4.43 MeV
=12*931 MeV +4.43 MeV
=11176.43 MeV
m(¹²N) =12.018613*931 MeV
=11189.329 MeV.
Q-value of the decay scheme,
=m(¹²N)-m(¹²C*) -2mₑ
=(11189.329 -11176.43) MeV -2*511 keV
=12.899 -1.022 MeV
=11.877 MeV
≈11.88 MeV.
This much energy will be shared by the ß-particle and the neutrino. For the maximum kinetic energy of the ß-particle, we take the energy of the neutrino =zero. The total of the above energy released will be taken by the ß-particle.
19. The decay constant of ¹⁹⁷₈₀Hg (electron capture to ¹⁹⁷₇₉Au) is 1.8x10⁻⁴ s⁻¹.
(a) What is the half-life?
(b) What is the average life?
(c) How much time will it take to convert 25% of this isotope of mercury into gold?
ANSWER: (a) Decay constant given,
λ =1.8x10⁻⁴ s⁻¹
Hence the half-life,
t½ =0.693/λ
=0.693/(1.8x10⁻⁴ s⁻¹)
=3850 s
≈64 min.
(b) Average life = 1/λ
=1/(1.8x10⁻⁴ s⁻¹)
=5555 s
≈92 min.
(c) 25% of the original isotope decays means only 75% of the original nuclei remain active after time t in the above decay.
N/Nₒ =3/4.
Let the time taken in this decay =t, then
N/Nₒ =e-λt =3/4
→-λt =ln (3/4)
→t = -0.2877/(-1.8x10⁻⁴ s⁻¹)
≈1600 s.
20. The half-life of ¹⁹⁸Au is 2.7 days. (a) Find the activity of a sample containing 1.0 µg of ¹⁹⁸Au. (b) What will be the activity after 7 days? Take the atomic weight of ¹⁹⁸Au to be 198 g/mol.
ANSWER: 198 g of this Au will contain a 6.02x10²³ number of atoms.
Number of atoms in 1.0µg,
N =1.0x10⁻⁶*6.02x10²³/198
=3.04x10¹⁵
Half-life (given) =2.7 days =2.354x10⁵ s
Hence decay constant,
λ =0.693/2.354x10⁵ s⁻¹
=2.944x10⁻⁶ s⁻¹
(a) Hence the activity of the sample,
Aₒ =λN
=2.944x10⁻⁶*3.04x10¹⁵ disintegration/s
=8.95x10⁹ Bq
=8.95x10⁹/3.7x10¹⁰ Ci
=0.242 Ci
(b) Activity at time t is,
A =Aₒ/2t/t½
=0.242/(27/2.7)
=0.040 Ci.
21. Radioactive ¹³¹I has a half-life of 8.0 days. A sample containing ¹³¹I has activity 20 µCi at t =0. (a) What is its activity at t =4.0 days? (b) What is its decay constant at t =4.0 days?
ANSWER: Given t½ =8 days
The decay constant λ =0.693/t½
=0.693/8 per day
=0.087 per day
Aₒ =20 µCi
(a) t =4.0 days
Activity at t =4.0 days
A =Aₒe-𝜆t
=(20 µCi)*e-(0.087*4.0)
=20*e-0.348 µCi
=14.12 µCi
≈14 µCi.
(b) The decay constant at t =4.0 days will remain the same as it depends only on the half-life of the sample. Hence,
λ =0.087 per day
=0.087/(24*3600) per second
=1.01x10⁻⁶ s⁻¹.
22. The decay constant of ²³⁸U is 4.9x10⁻¹⁸ s⁻¹. (a) What is the average life of ²³⁸U? (b) What is the half-life of ²³⁸U? (c) By what factor does the activity of a ²³⁸U sample decrease in 9x10⁹ years?
ANSWER: Given the decay constant,
λ =4.9x10⁻¹⁸ s⁻¹
(a) Average life of the given sample of U,
tₐᵥ =1/λ
=1/4.9x10⁻¹⁸ s
=2.041x10¹⁷ s
=6.47x10⁹ years.
(b) Half-life of the given sample,
t½ =0.693/λ
=0.693/4.9x10⁻¹⁸ s
=1.414x10¹⁷ s
=4.5x10⁹ y.
(c) Aₒ/A =2t/t½
Here t/t½ =9x10⁹/4.5x10⁹ =2
Hence Aₒ/A =2² =4.
23. A certain sample of a radioactive material decays at the rate of 500 per second at a certain time. The count rate falls to 200 per second after 50 minutes. (a) What is the decay constant of the sample? (b) What is its half-life?
ANSWER: Given Aₒ =500 per second.
After time t =50 min =3000 s,
Remaining activity,
A =200 per second.
(a) The decay constant lambda is given as,
A =Aₒ*e-λt
→λt =ln (Aₒ/A) =ln (500/200) =0.916
→λ =0.916/3000 per second
=3.05x10⁻⁴ s⁻¹.
(b) Half-life =1/λ
=1/3.05x10⁻⁴ s
=2272 s ≈ 38 min.
24. The count rate from a radioactive sample falls from 4.0x10⁶ per second to 1.0x10⁶ per second in 20 hours. What will be the count rate 100 hours after the beginning?
ANSWER: eλt =Aₒ/A
λt =ln (Aₒ/A)
→λ ={ln (4x10⁶/1x10⁶)}/t
={ln 4}/20 per hr
=0.0693 per hr
If we take t =100 hours from the beginning,
λt =0.0693*100 =6.93
Hence from,
eλt =Aₒ/A
A =Aₒ*e-6.93
=4x10⁶*e-6.93 per second
=3.9x10³ per second.
25. The half-life of ²²⁶Ra is 1602 y. Calculate the activity of 0.1 g of RaCl₂ in which all the radium is in the form of ²²⁶Ra. Taken atomic weight of Ra is 226 g/mol and that of Cl is 35.5 g/mol.
ANSWER: Molecular weight of RaCl₂
=226+2*35.5 =297
Hence 0.1 g of RaCl₂ will contain molecules,
N =0.1*6.02x10²³/297 molecules
=2.03x10²⁰ molecules.
Half-life t½ =1602 y
Decay constant
λ =0.693/1602 per year
=4.33x10⁻⁴ /y
Activity A =λN
=4.33x10⁻⁴*2.03x10²⁰/(365*24*3600) s⁻¹
=2.8x10⁹ disintegrations/s.
26. The half-life of a radioisotope is 10 h. Find the total number of disintegrations in the tenth hour measured from a time when the activity was 1 Ci.
ANSWER: Given t½ =10 h. Hence the decay constant,
𝜆 =0.693/10 =0.0693 per h.
=1.925x10⁻⁵ s⁻¹
Given Aₒ =1 Ci =3.7x10¹⁰ disintegrations/s.
Number of active nuclei at t =0,
Nₒ =Aₒ/𝜆
=3.7x10¹⁰/1.925x10⁻⁵
=1.922x10¹⁵
Number of active nuclei at t =9 h,
N =Nₒ e-𝜆t
=1.922x10¹⁵*e-1.925x10⁻⁵*9*3600
=1.030x10¹⁵
Since the half-life of the sample is 10 h, the number of active nuclei at t =10h,
N½ =Nₒ/2
=1.922x10¹⁵/2
=0.961x10¹⁵
Hence the total number of disintegrations in the tenth hour,
=N -N½
=(1.030 -0.961)x10¹⁵
=6.9x10¹³.
27. The selling rate of a radioactive isotope is decided by its activity. What will be the second-hand rate of a one-month-old ³²P (t½ =14.3 days) source if it was originally purchased for 800 rupees?
ANSWER: Given t½ =14.3 days.
Ratio of the original activity remaining,
A/Aₒ =1/2t/t½
=1/2(30/14.3) =0.234
Since the rate is proportional to its activity, the second-hand rate after one month will be reduced in the above proportion.
This rate is =0.234x800 rupees
≈187 rupees.
28. ⁵⁷Co decays to ⁵⁷Fe by ß⁺-emission. The resulting ⁵⁷Fe is in its excited state and comes to the ground state by emitting 𝛾-rays. The half-life of ß⁺-decay is 270 days and that of 𝛾-emission is 10⁻⁸ s. A sample of ⁵⁷Co gives 5.0x10⁹ gamma rays per second. How much time will elapse before the emission rate of gamma rays drops to 2.5x10⁹ per second?
ANSWER: Since the gamma decay from an active nucleus is only after a beta-decay, the number of gamma rays will be equal to the beta-decays. Since the number of beta decay is half after 270 days, the number of gamma rays will also drop to half after 270 days.
29. Carbon (Z =6) with mass number 11 decays to boron (Z =5). (a) Is it a ß⁺-decay or ß⁻-decay? (b) The half-life of the decay scheme is 20.3 minutes. How much time will elapse before a mixture of 90% carbon-11 and 10% boron-11 (by the number of atoms) converts itself into a mixture of 10% carbon-11 and 90% boron-11?
ANSWER: (a) As a result of decay, the atomic number decreases which means the number of protons in a decayed nucleus has reduced by one. Clearly, a proton has changed to a neutron and a positron. The positron is emitted as ß⁺-decay. So it is a ß⁺-decay.
(b) t½ =20.3 minutes =1218 s.
Suppose we have a sample of this mixture with a total of 100 atoms. Out of this, only 90 atoms of carbon-11 will decay with time. Let after time t, 80 atoms of carbon-11 decay to boron-11. Now the mixture has 10 carbon-11 atoms and 90 boron-11 atoms.
Here Nₒ =90, N =10,
𝜆 =0.693/t½
=0.693/1218 per second
=5.69x10⁻⁴ s⁻
Since N =Nₒ*e-𝜆t
→𝜆t = ln (Nₒ/N) =ln (90/10)
→t =2.197/(5.69x10⁻⁴) s
=3861 s
=64 min.
30. 4x10²³ tritium atoms are contained in a vessel. The half-life of decay of tritium nuclei is 12.3 y. Find (a) the activity of the sample, (b) the number of decays in the next 10 hours, and (c) the number of decays in the next 6.15 y.
ANSWER: (a) t½ =12.3 y
=12.3*365*24*3600 s
=3.88x10⁸ s
Decay constant 𝜆 =0.693/t½
=1.787x10⁻⁹ s⁻
The activity of the sample,
Aₒ =𝜆Nₒ
=1.787x10⁻⁹*4x10²³ disintergrations/s
=7.146x10¹⁴ disintegrations/s.
(b) t =10 hours =36000 s
𝜆t =1.787x10⁻⁹*36000 =6.433x10⁻⁵
Number of active nuclei after 10 hours,
N =Nₒ*e⁻𝜆t
=4x10²³*e-6.433x10⁻⁵
=3.999742x10²³
Hence the number of decays in 10 hours,
=Nₒ -N
=4x10²³ -3.999742x10²³
=2.57x10¹⁹.
(c) t =6.15 y =1.94x10⁸ s
𝜆t =1.787x10⁻⁹*1.94x10⁸ =0.347
N =Nₒ*e-𝜆t
=4x10²³*e-0.347
=2.83x10²³
The number of decays in this period,
=4x10²³ -2.83x10²³
=1.17x10²³.
31. A point source emitting alpha particles is placed at a distance of 1 m from a counter which records any alpha particle falling on its 1 cm² window. If the source contains 6.0x10¹⁶ active nuclei and the counter records a rate of 50,000 counts per second, find the decay constant. Assume that the source emits alpha particles uniformly in all directions and that the alpha particles fall nearly normally on the window.
ANSWER: Surface area of a sphere of radius 1 m with the point source at the center,
=4πr²
=4π*1² m²
=4π m²
Number of alpha particles falling on 1 cm²,
=50000 counts/s.
Total alpha particles emitted by the point source =Total counts on the surface area of the above sphere,
=4π*10000*50000 counts/s
=2x10⁹π counts per second.
So the activity of sample A =2x10⁹π per second.
A =𝜆N
→𝜆 =A/N
=2x10⁹π/6x10¹⁶ per s
=1.05x10⁻⁷ per s
Hence the decay constant, 𝜆 =1.05x10⁻⁷ s⁻.
32. ²³⁸U decays to ²⁰⁶Pb with a half-life of 4.47x10⁹ y. This happens in a number of steps. Can you justify a single half-life for this chain of processes? A sample of rock is found to contain 2.00 mg of ²³⁸U and 0.600 mg of ²⁰⁶Pb. Assuming that all the lead has come from uranium, find the life of the rock.
ANSWER: The unstable nucleus of ²³⁸U decays to different unstable nuclei in a number of steps and finally decays to the stable nucleus of ²⁰⁶Pb, each step having its own half-life. But this final change of nucleus from U to Pb will have a fixed decay constant and hence a single half-life for this chain of processes can be justified.
238 g of ²³⁸U and 206 g of ²⁰⁶Pb each contain 6.022x10²³ number of atoms.
0.600 mg of Pb will contain,
0.6*6.022x10²³/(206*1000) atoms
=1.75x10¹⁸ atoms.
Similarly, 2.00 mg of the given uranium will contain,
2*6.022x10²³/(238*1000) atoms
=5.06x10¹⁸ atoms.
Hence the original number of uranium atoms,
Nₒ =(1.75+5.06)x10¹⁸ atoms
=6.81x10¹⁸ atoms.
The present number of uranium atoms,
N =5.06x10¹⁸ atoms. If 𝜆 is the decay constant then,
N =Nₒe-𝜆t
5.06x10¹⁸ =6.81x10¹⁸*e(-0.693/4.47x10⁹)t,
{where t is the life of the rock}
→e(-1.55x10⁻¹⁰)t =5.06/6.81
→t ={ln (5.06/6.81)}/(-1.55x10⁻¹⁰) y.
=1.92x10⁹ y.
33. When charcoal is prepared from a living tree, it shows a disintegration rate of 15.3 disintegrations of ¹⁴C per gram per minute. A sample from an ancient piece of charcoal shows ¹⁴C activity to be 12.3 disintegrations per gram per minute. How old is this sample? The half-life of ¹⁴C is 5730 y.
ANSWER: The present activity of an ancient sample of charcoal,
A =12.3 disintegrations/minute.
Fresh charcoal's activity,
Aₒ =15.3 disintegrations/minute
The half-life of charcoal
t½ =5730 y
Decay constant 𝜆 =0.693/5730 per y
=1.20x10⁻⁴ per y
Suppose the age of the sample =t years
A =Aₒe⁻𝜆t
e⁻𝜆t =A/Aₒ
→⁻𝜆t =ln (12.3/15.3)
→t =-0.218/(-1.20x10⁻⁴)
≈1800 y.
34. Natural water contains a small amount of tritium (³₁H). This isotope beta-decays with a half-life of 12.5 years. A mountaineer while climbing towards a difficult peak finds debris of some earlier unsuccessful attempt. Among other things, he finds a sealed bottle of whiskey. On return, he analyses the whiskey and finds that it contains only 1.5 percent of ³₁H radioactivity as compared to a recently purchased bottle marked '8 years old'. Estimate the time of that unsuccessful attempt.
ANSWER: Given A/Aₒ =1.5% =0.015
But A/Aₒ =e⁻𝜆t
Hence e⁻𝜆t =0.015
→⁻𝜆t =ln (0.015) =-4.2
→t =4.2/𝜆
=4.2*t½/0.693
=6.06*12.5 y
=75.7 y
But the recently purchased whisky was manufactured 8 years ago. So the above duration is up to that time. Hence from now on, the unsuccessful event was 75.7+8 =83.7 y, hence about 84 y ago.
35. The count rate of nuclear radiation coming from a radioactive sample containing ¹²⁸I varies with time as follows,
Time t (minutes); 0 25 50 75 100
Count rate R(10⁹ s⁻¹): 30 16 8.0 3.8 2.0
(a) Plot ln(Rₒ/R) against t. (b) From the slope of the best straight line through the points, find the decay constant 𝜆. (c) Calculate the half-life t½.
ANSWER: (a) Given Rₒ =30x10⁹ per s
Hence the variations of ln (Rₒ/R) with time in the given problem will be as follows:
Time (minutes) : 0 25 50 75 100
Rₒ/R : 1 1.88 3.75 7.89 15
ln (Rₒ/R) : 0 0.63 1.32 2.07 2.71
It can be plotted as follows;
 |
| Plot for Q-35(a) |
(b) The best line is drawn through the points and the slope of this line is,
m =AB/BO
=2.70/100 per min
=0.027 per min.
Since ln(Rₒ/R) =𝜆t, that is in the form of y =mx, a straight line passing through the origin. The slope of the line will give the value of the decay constant λ.
Hence 𝜆 =m =0.027 per min.
(c) Hence half-life,
t½ =0.693/𝜆
=0.693/0.027 min
≈ 25 min.
36. The half-life of ⁴⁰K is 1.3x10⁹ y. A sample of 1 g of pure KCl gives 160 counts/s. Calculate the relative abundance of ⁴⁰K (fraction of ⁴⁰K present) in natural potassium.
ANSWER: The average atomic mass of K =39. Hence 39+35.5 =74.5 g of KCl will have 6.022x10²³ potassium atoms.
Hence 1 g of KCl will have 6.022x10²³/74.5 atoms,
N'=8.08x10²¹ atoms.
Given that count rate A =160 counts/s and t½ =1.3x10⁹ y
Decay constant 𝜆 =0.693/1.3x10⁹ per y
=5.33x10⁻¹⁰ per y
The number N of active atoms of ⁴⁰K is given as
A =𝜆N
→N =A/𝜆
=160*3600*24*365/5.33x10⁻¹⁰
=9.465x10¹⁸
Hence the relative abundance of ⁴⁰K present in the sample is,
N/N' =9.465x10¹⁸/8.08x10²¹
=0.0012
=0.12%.
37. ¹⁹⁷₈₀Hg decays to ¹⁹⁷₇₉Au through electron capture with a decay constant of 0.257 per day. (a) What other particle/particles are emitted in the decay? (b) Assume that the electron is captured from the K-shell. Use Moseley's law √v =a(Z-b) with a =4.95x10⁷ s⁻¹⁾² and b =1 to find the wavelength of the Kₐ X-ray emitted following the electron capture.
ANSWER: (a) In the decay by electron capture, the nucleus captures one of its orbiting electrons and one of the protons converts into a neutron. The atomic number of the new nucleus is reduced by one and a new element is formed. In this process, a neutrino is emitted from the nucleus.
(b) Assuming that an electron is captured from the K-shell, the frequency of the radiation of Kₐ X-ray is given from Moseley's law,
√𝜈 =a(Z -b)
=4.95x10⁷*(79 -1)
→𝜈 =1.49x10¹⁹
Hence the wavelength,
=c/𝜈
=3x10⁸/1.49x10¹⁹ m
=2.0x10⁻¹¹ m
=20x10⁻¹² m
=20 pm.
38. A radioactive isotope is being produced at a constant rate dN/dt =R in an experiment. The isotope has a half-life t½. Show that after a time t >> t½ the number of active nuclei will become constant.
ANSWER: The rate of production of radioactive isotope =R =constant.
In a very small time dt, isotopes produced =Rdt. If at t =0, there are N existing isotopes, then after time t remaining active nuclei.
=Ne-𝜆t +∫(Rdt)e-𝜆t
=Ne-𝜆t +R∫e-𝜆t dt,
{Between limits t =0 to t =t}
=Ne-𝜆t +(-R/𝜆)[e-𝜆t]₀t
=Ne-𝜆t +(R/𝜆)(1-e-𝜆t)
=R/𝜆 +(N -R/𝜆)e-𝜆t
If t >> t½, then e-𝜆t →0.
In this case, the number of active nuclei will be ≈R/𝜆.
=R*t½/0.693.
It is a constant.
39. Consider the situation of the previous problem. Suppose the production of the radioactive isotope starts at t =0. Find the number of active nuclei at time t.
ANSWER: Consider a very small interval of time dt just after t =0. The number of nuclei produced in this time =Rdt.
We should keep in mind that as soon as radioactive isotopes are produced, they start decaying with their own decay constant. Hence after time t, the remaining active nuclei of Rdt will be
=(Rdt)e-𝜆t
Since the nuclei are being produced continuously, The number of active nuclei after time t
=∫(Rdt)e-𝜆t
=R∫e-𝜆t dt,
{Between limits t =0 to t =t}
=(-R/𝜆)[e-𝜆t]₀t
=(-R/𝜆)(e-𝜆t -1)
=(R/𝜆)(1 -e-𝜆t).
40. In an agricultural experiment, a solution containing 1 mole of radioactive material (t½ =14.3 days) was injected into the roots of a plant. The plant was allowed 70 hours to settle down and then activity was measured in its fruits. If the activity measured was 1 µCi, what percent of activity is transmitted from the root to the fruit in a steady state?
ANSWER: One mole of active nuclei means the number of nuclei,
=6.022x10²³.
After 70 days (=2.9 days), the remaining active nuclei,
N =Nₒ/(2t/t½)
=6.022x10²³/(20.2)
=5.24x10²³
𝜆 =0.693/t½
=0.693/14.3
=0.048 per day
Activity A =𝜆N
→A =0.048*5.24x10²³ per day
=2.52x10²² per day
=2.91x10¹⁷ per second
Fruits get 1µCi of nuclei which is equal to,
=3.7x10¹⁰*10⁻⁶
=3.7x10⁴ count per second.
Hence the percentage of activity transferred from root to fruits in a steady state,
=3.7x10⁴*100/2.91x10¹⁷
=1.27x10⁻¹¹%.
41. A vessel of volume 125 cm³ contains tritium (³H, t½ =12.3 y)at 500 kPa and 300 K. Calculate the activity of the gas.
ANSWER: P =500 kPa =5x10⁵ Pa,
V =125 cm³ =1.25x10⁻⁴ m³, T =300 K.
Gas constant R =8.314 J/(mol.K).
Hence the number of moles in the vessel,
n =PV/RT
=5x10⁵*1.25x10⁻⁴/(8.314*300) mol
=0.025 mol
Hence the number of tritium nuclei,
N =0.025*6.022x10²³
=1.50x10²².
Half-life t½ =12.3 y
Decay constant 𝜆 =0.693/t½
→𝜆 =0.693/12.3 per y
=0.0563 /y
Activity =𝜆N
=0.0563*1.50x10²²/(365*24*3600) dis/s
=2.68x10¹³ disintegrations/s
=724 Ci.
42. ²¹²₈₃Bi can disintegrate either by emitting an α-particle or by emitting a ß⁻-particle. (a) Write the two equations showing the products of the decays. (b) The probabilities of disintegration by α- and ß- decays are in the ratio 7:13. The overall half-life of ²¹²Bi is one hour. If 1 g of pure ²¹²Bi is taken at 12:00 noon, what will be the composition of this sample at 1 p.m. the same day?
ANSWER: (a) When the given isotope of Bi decays by emitting an alpha particle, two neutrons, and two protons in the parent nucleus are reduced because the emitted alpha particle is a helium nucleus. The equation is as follows;
²¹²₈₃Bi → ²⁰⁸₈₁Tl +⁴₂He.
When one of the neutrons in the nucleus converts into a proton and an electron, the electron is emitted as a beta particle along with an anti-neutrino. The number of protons in the nucleus increases by one. For the decay by emitting a ß⁻-particle, the equation may be written as follows;
²¹²₈₃Bi →²¹²₈₄Po +e⁻ +𝜈̅
(b) Since the half-life of Bi is one hour, at 1 pm only 0.5 g of Bi will remain in the sample.
By the given probability of the type of decay,
0.5*7/20 g =0.175 g will be decayed by the emission of alpha particles. So approximately 0.175 g will be Tl.
Decay by emission of beta particles will be for 0.5*13/20 g =0.325 g of Bi. So approximately 0.325 g of Po will be present in the sample.
43. A sample contains a mixture of ¹⁰⁸Ag and ¹¹⁰Ag isotopes each having an activity of 8.0x10⁸ disintegrations per second. ¹¹⁰Ag is known to have a larger half-life than ¹⁰⁸Ag. The activity A is measured as a function of time and the following data are obtained. 
(a) Plot ln(A/Aₒ) versus time. (b) See that for large values of time, the plot is nearly linear. Dedude the half-life of ¹¹⁰Ag from this portion of the plot. (c) Use the half-life of ¹¹⁰Ag to calculate the activity corresponding to the ¹⁰⁸Ag in the first 50 s. (d) Plot ln(A/Aₒ) versus time for ¹⁰⁸Ag for the first 50 s. (e) Find the half-life of ¹⁰⁸Ag.
ANSWER: (a) Activity at time t =0 is,
Aₒ =2*8x10⁸ disintegrations per second,
=16x10⁸ disintegrations per second.
From the given chart we calculate,
Time(s) (A/Aₒ) ln(A/Aₒ)
0 1 0
20 0.737 -0.30
40 0.573 -0.56
60 0.466 -0.76
80 0.392 -0.94
100 0.338 -1.08
200 0.193 -1.65
300 0.118 -2.14
400 0.073 -2.62
500 0.045 -3.10
ln(A/Aₒ) vs. time may be plotted as follows:-
 |
| Plot for t =0 to t =100 s |
For t =100 s to t =500 s we plot as foloows,
 |
| Plot for t =100 s to t =500 s |
(b) As we see in the second plot, for a large value of time the plot is nearly linear. This is due to the fact that this graph mostly represents the activity of the isotope ¹¹⁰Ag because it has a longer half-life. Since
ln (A/Aₒ) =-𝜆t
Which is a graph of a straight line in the form of y =mx. Here the slope of the graph gives the value of decay constant 𝜆. The slope of the graph from the second plot above,
𝜆 =1.45/300 s⁻¹
=0.0048 s⁻¹
Hence the half-life of ¹¹⁰Ag,
=0.693/0.0048 =144 s.
(c) Activity of ¹¹⁰Ag in first 50 s
A =Aₒ/2(50/144)
=8x10⁸/1.272
=6.29x10⁸ disintegrations per second.
From the first graph, for t =50 s,
ln (A/Aₒ) =-0.66
→A =16x10⁸*0.517
=8.27x10⁸
Hence the activity of ¹⁰⁸Ag after the first 50 s
=8.27x10⁸ -6.29x10⁸
=1.98x10⁸ disintegrations /s
(d) Time Activity of Activity of ln(A/Aₒ)
(s) ¹¹⁰Ag (*10⁸) ¹⁰⁸Ag (*10⁸) of ¹⁰⁸Ag
20 7.26 4.54 -0.57
40 6.60 2.57 -1.09
50 6.29 1.98 -1.40
The graph for ln(A/Aₒ) versus Time (for the first 50 s) may be plotted as follows:--
 |
| Plot for t =0 to t =50 s for ¹⁰⁸Ag |
(e) Decay constant =slope of the graph,
→𝜆 =1.38/50 s⁻¹
=0.028 s⁻¹
Half-life t½ =0.693/0.028 s
=24.7 s.
44. A human body excretes (removes by waste discharge, sweating, etc.) certain materials by a law similar to radioactivity. If technetium is injected in some form into a human body, the body excretes half the amount in 24 hours. A patient is given an injection containing ⁹⁹Tc. This isotope is radioactive with a half-life of 6 hours. The activity from the body just after the injection is 6 µCi. How much time will elapse before the activity falls to 3 µCi?
ANSWER: Given, that Aₒ =6 µCi, A =3 µCi,
t½ =6 hours, suppose after time t the above change in activity takes place.
Since t½ for excreting the isotope =24 hours. After time t, the remaining nuclei,
N =Nₒ e-𝜆t
But in this time remaining active nuclei out of N due to disintegration,
N' =N e-𝜆't =Nₒe-𝜆t*e-𝜆't
→N' =Nₒe-(𝜆+𝜆')t
→e-(𝜆+𝜆')t =N'/Nₒ =A'/Aₒ =1/2
→-(𝜆+𝜆')t =ln(1/2) =-0.693
→(0.693/t½ +0.693/t'½)t =0.693
→t =t½*t'½/(t½ +t'½)
=6*24/(6 +24)
=4.8 hours.
45. A charged capacitor of capacitance C is discharged through a resistance R. A radioactive sample decays with an average-life 𝜏. Find the value of R for which the ratio of the electrostatic field energy stored in the capacitor to the activity of the radioactive sample remains constant in time.
ANSWER: In a discharging circuit, the charge on the capacitor at time t is given as,
q =Qe-t/RC
Hence the electrostatic energy stored in the capacitor at time t is,
U =½q²/C
=½(Q²e-2t/RC)/C
The average life of the radioactive sample =𝜏.
Hence its decay constant, 𝜆 =1/𝜏
The activity of the sample at time t,
A =Aₒ*e-𝜆t
=Aₒe-t/𝜏
Given that U/A =constant.
→(½Q²e-2t/RC/C)/(Aₒe-t/𝜏) =K (say)
→(½Q²/C)e-2t/RC =(KAₒ)e-t/𝜏
Comparing and equating the powers of e on both sides, we get
-2t/RC =-t/𝜏
→RC/2 =𝜏
→R =2𝜏/C.
46. Radioactive isotopes are produced in a nuclear physics experiment at a constant rate dN/dt =R. An inductor of inductance 100 mH, a resistor of resistance 100 Ω, and a battery are connected to form a series circuit. The circuit is switched on at the instant the production of radioactive isotope starts. It is found that i/N remains constant in time where i is the current in the circuit at time t and N is the number of active nuclei at time t. Find the half-life of the isotope.
ANSWER: L =100 mH, r =100 Ω. Let emf of the battery = E. The current in this circuit at time t,
i =(E/r)(1 -e-tr/L).
The number of radioactive isotopes produced in time dt just after the production starts =Rdt. It starts decaying with its own decay constant.
The number of active nuclei after time t,
=Rdt*e-𝜆t
Integrating it within time limits zero to t we get,
The number of active nuclei after time t,
N =(R/𝜆)(1 -e-𝜆t)
{see integration in the solution of problem number 39 of this chapter → Link of the post}
Given that i/N =constant =K (say). So,
(E/r)(1 -e-tr/L)/{(R/𝜆)(1 -e-𝜆t)} =K
→(E/r)(1 -e-tr/L) =(KR/𝜆)(1 -e-𝜆t)
Comparing both sides and equating the powers of e, we get
-tr/L =-𝜆t
→𝜆 =r/L
The half-life of the isotope,
=0.693/𝜆
=0.693L/r
=0.693*(100/1000)/100 s
=0.693/1000 s
=6.93x10⁻⁴ s.
47. Calculate the energy released by 1 g of natural uranium assuming 200 MeV is released in each fission event and that the fissionable isotope ²³⁵U has an abundance of 0.7% by weight in natural uranium.
ANSWER: The mass of fissionable isotope ²³⁵U in 1 g of natural uranium,
m =(0.7/100)*1 g
=7x10⁻³ g
The number of nuclei in this mass,
=6.022x10²³*(7x10⁻³/235)
=1.794x10¹⁹
Hence the energy released by this number of fissionable nuclei
E =1.794x10¹⁹*200 MeV
=3.59x10²¹ MeV
=3.59x10²¹*1.602x10⁻¹⁹*10⁶ J
=5.75x10⁸ J.
48. A uranium reactor develops thermal energy at a rate of 300 MW. Calculate the amount of ²³⁵U being consumed every second. The average energy released per fission is 200 MeV.
ANSWER: Thermal energy is produced at a rate of 300 MW =300 MJ/s.
Per second energy produced,
E =300x10⁶ J
=3x10⁸ J
=3x10⁸/(1.602x10⁻¹⁹) eV
=1.87x10²⁷ eV
=1.87x10²¹ MeV.
The number of nuclei consumed per second
N =9.35x10¹⁸
235 g of ²³⁵U contains 6.022x10²³ nuclei.
Hence 9.35x10¹⁸ nuclei will have a mass of
m =(9.35x10¹⁸/6.022x10²³)*235 g
=0.00365 g
≈3.7 mg.
Thus the amount of ²³⁵U consumed per second in the reactor =3.7 mg.
49. A town has a population of 1 million. The average electric power needed per person is 300 W. A reactor is to be designed to supply power to this town. The efficiency with which thermal power is converted into electric power is aimed at 25%. (a) Assuming 200 MeV of thermal energy to come from each fission event on average, find the number of events that should take place every day. (b) Assuming the fission to take place largely through ²³⁵U, at what rate will the amount of ²³⁵U decrease? Express your answer in kg/day. (c) Assuming that uranium enriched to 3% in ²³⁵U will be used, how much uranium is needed per month (30 days)?
ANSWER: Electric power needed for the town,
=300*10⁶ W
=3x10⁸ J/s
=3x10⁸/(1.602x10⁻¹⁹*10⁶) MeV/s
=1.87x10²¹ MeV/s.
Assuming the efficiency with which thermal power is converted into electric power is 25%, thermal power needed per second,
=1.87x10²¹/0.25 MeV
=7.48x10²¹ MeV.
For the whole day power is needed,
=7.48x10²¹*24*3600 MeV
=6.46x10²⁶ MeV.
(a) Each fission event releases 200 MeV of energy, hence the number of events taking place each day,
=6.46x10²⁶/200
=3.23x10²⁴.
(b) 6.022x10²³ number of atoms make 235 g of ²³⁵U, hence 3.23x10²⁴ atoms of ²³⁵U will make
(3.23x10²⁴/6.022x10²³)*235 g
=1261 g
=1.261 kg
Hence 1.261 kg/day of ²³⁵U will decrease each day.
(c) If 3% ²³⁵U enriched uranium is to be used then the per day requirement of uranium,
=1.261/(0.03) kg
=42.033 kg
Requirement of uranium per month,
=30*42.033 kg
=1261 kg.
50. Calculate the Q-values of the following fusion reactions:
(a) ²₁H + ²₁H → ³₁H +¹₁H
(b) ²₁H + ²₁H → ³₂He + n
(c) ²₁H + ³₁H → ⁴₂He + n.
Atomic masses are m(²₁H) =2.014102 u, m(³₁H) =3.016049 u, m(³₂He) =3.016029 u, and m(⁴₂He) =4.002603 u.
ANSWER: (a) Uᵢ =2*m(²₁H) -2mₑ
=2*2.014102 -2*0.0005486 u
=4.0271068 u
Uf =m(³₁H) +m(¹₁H) -2mₑ
=3.016049 +1.007825 -2*0.0005486 u
=4.0227768 u
Q-value =Uᵢ -Uf
=4.0271068 -4.0227768 u
=0.00433 u
=0.00433*931 MeV.
=4.03 MeV.
(b) Uᵢ =2*m(²₁H) -2mₑ
Uf =m(³₂He) +mₙ -2mₑ
Q =2*m(²₁H) -m(³₂He) -mₙ
=2*2.014102 -3.016029 -1.008665 u
=0.00351 u
=0.00351*931 MeV
=3.26 MeV.
(c) Q =m(²₁H) +m(³₁H) -m(⁴₂He) -mₙ
=2.014102 +3.016049 -4.002603 -1.008665 u
=0.01888 u
=0.01888*931 MeV
=17.57 MeV.
51. Consider the fusion in helium plasma. Find the temperature at which the average thermal energy 1.5 kT equals the coulomb potential energy at 2 fm.
ANSWER: Charge on each helium nucleus,
q =2e.
Coulomb potential energy between two helium nuclei at a distance of r =2 fm,
U =(1/4π𝜀ₒ)(2e)²/r
=(9x10⁹ N-m²/C²)(2*1.6x10⁻¹⁹ C)²/(2x10⁻¹⁵ m)
=4.61x10⁻¹³ J.
Average thermal energy =1.5 kT, where k is the Boltzmann constant.
Given that, 1.5 kT =U
→T =U/1.5k
=(4.61x10⁻¹³ J)/(1.5*1.38x10⁻²³ J/K)
=2.23x10¹⁰ K.
52. Calculate the Q-value of the fusion reaction
⁴He + ⁴He =⁸Be.
Is such a fusion energetically favorable? The atomic mass of ⁸Be is 8.0053 u and that of ⁴He is 4.0026 u.
ANSWER: The Q-value of the given fusion reaction is,
Q =2*m(⁴He) -m(⁸Be)
=2*4.0026 -8.0053 u
=-0.0001 u
=-0.0001*931 MeV
=-0.0931 MeV
=-93.1 KeV.
Since the Q-value is negative, this fusion reaction will not release energy but require energy. So it is not energetically favorable.
53. Calculate the energy that can be obtained from 1 kg of water through the fusion reaction
²H +²H → ³H + p.
Assume that 1.5x10⁻²% of natural water is heavy water D₂O (by the number of molecules) and all the deuterium is used for fusion.
ANSWER: Q-value of the reaction,
Q ={2*m(²H) -2mₑ} -{m(³H) -mₑ +mₚ}
=2m(²H) -m(³H) -mₑ -mₚ
=2*2.014102 -3.016029 -0.0005486 -1.007276 u
=0.00435 u
=4.05 MeV
18 g of natural water contains 6.022x10²³ molecules. So 1000 g (1 kg) will contain
(1000/18)*6.022x10²³ molecules,
=3.3456x10²⁵ molecules.
From the given percentage of heavy water, the number of D₂O molecules,
=(1.5x10⁻²/100)*3.3456x10²⁵ molecules
=5.0184x10²¹ molecules.
Each molecule of the heavy water contains two deuterium atoms, hence a total of 5.0184x10²¹ fusion reaction will take place if all the deuterium is to be consumed. So the total energy obtained is,
=5.01x10²¹*4.05 MeV
=2.029x10²²*1.6x10⁻¹⁹ MJ
=3246 MJ.
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