Questions for Short Answer
1. When a Coolidge tube is operated for some time it becomes hot. Where does the heat come from?
ANSWER: In the Coolidge tube, electrons from the hot filament are accelerated to hit the metal target to emit the X-rays but not all the electrons are able to emit X-rays. Some electrons collide with metal atoms and lose their kinetic energy. The kinetic energy of these absorbed electrons is utilized to increase the vibration of the metal molecules. It increases the temperature of the metal. So the heat in the Coolidge tube comes from the conversion of the kinetic energy of the electrons.
2. In a Coolidge tube, electrons strike the target and stop inside it. Does the target get more and more negatively charged as time passes?
ANSWER: The target is not an isolated piece of metal but connected to the positive end of a battery terminal to create a large potential difference between the hot filament and the target to accelerate the electrons emitted from the filament. Thus, those electrons that stop inside the target do not remain inside it but go through the outer circuit into the battery. So the target does not get more and more negatively charged with time.
3. Can X-rays be used for the photoelectric effect?
ANSWER: Of course. If the energy of the X-ray photon is more than the threshold frequency of the metal it can eject photoelectrons from that metal surface.
4. Can X-rays be polarized?
ANSWER: Since X-rays are also electromagnetic waves, they are transverse in nature. Since transverse waves can be polarised, X-rays can also be polarised. If we consider the particle nature of the light that consists of photons, the polarization can be explained through quantum theory.
5. X-ray and visible light travel at the same speed in a vacuum. Do they travel at the same speed in the glass?
ANSWER: The refractive index of glass for X-rays is less than that for visible light. Since the speed of an electromagnetic wave is inversely proportional to the refractive index, the speed of X-rays in the glass will be more than the visible light.
6. Characteristic X-rays may be used to identify the element from which they are coming. Can continuous X-rays be used for this purpose?
ANSWER: Characteristic X-rays are produced when the accelerated electrons knock out an inner electron from the atom of the target. This vacancy is filled by the higher-orbit electrons in the atom. The difference in the energy levels of these two orbits is given to the emitted photon of the X-ray. This difference is characteristic of a particular element. By studying these characteristic X-rays the element can be identified from which they are emitted.
Continuous X-rays are emitted when the accelerated electrons lose their kinetic energy after hitting the target and stop inside the target after many collisions. So the energy (and frequency) of the photons thus emitted depends upon the amount of lost kinetic energy of electrons. It does not carry any information about the elements from which they are emitted. So continuous X-rays can not be used for identifying elements from which they are emitted.
7. Is it possible that in a Coolidge tube characteristic Lₐ X -rays are emitted but not Kₐ rays?
ANSWER: Kₐ X-ray is emitted when an electron in the K shell of the target atom is knocked out by the accelerating electron in the Coolidge tube and the electron in the L shell fills it. But the Lₐ X-ray is emitted if an electron in the L shell is knocked out and it is filled by an M shell electron in the target atom. We know that it requires less energy to knock out an electron from the L shell than the K shell. If the potential difference in the X-ray tube is maintained such that it gives the accelerating electrons less energy than required for dislodging a K shell electron but more energy than required to dislodge an electron from the L shell, then Lₐ X-rays may be emitted but not Kₐ X-rays. Hence in a Coolidge tube, it is possible that only Lₐ X-rays or other L series X-rays are emitted and not the Kₐ X-rays by controlling the potential difference in the tube.
8. Can Lₐ X-ray of one material have a shorter wavelength than Kₐ X-ray of another?
ANSWER: In other words, can an Lₐ X-ray photon of one material have more energy than a Kₐ X-ray photon of another material. The energy levels of K, L, and M shells depend on the atomic number Z of an element. The energy of the photon of Lₐ X-ray is the difference between the L and M shell of the atom while the energy of the photon of Kₐ X-ray is the difference between the K shell and L shell of the atom. There may be a situation in which an element with a higher atomic number such as Z =74 (Tungston) has an energy difference between the L and M shell more than the energy difference between the K and L shell of an element with a lower atomic number say Z =27 (Cobalt). In such a case, Lₐ X-rays of Tungston will have a shorter wavelength than the Kₐ X-rays of Cobalt.
9. Can a hydrogen atom emit characteristic X-rays?
ANSWER: The maximum energy of radiation that may be emitted from a hydrogen atom is 13.6 eV while the minimum energy of X-rays is 100 eV. So a hydrogen atom can not emit characteristic X-rays.
10. Why is exposure to X-ray injurious to health but exposure to visible light is not, when both are electromagnetic waves?
ANSWER: It is due to the fact that the photons of X-rays have much more energy than the photons of visible light. With very high frequency, X-rays have much penetrating power with damaging effects on living cells. Long exposures may cause cancer, alter the DNA of cells, and genetic defects.
OBJECTIVE-I
1. X-ray beam can be deflected
(a) by an electric field
(b) by a magnetic field
(c) by an electric as well as a magnetic field
(d) neither by an electric field nor by a magnetic field.
Answer: (d).
Explanation: X-rays are not charged particles, they are transverse electromagnetic radiation waves. These waves have a dual nature. Even when explained through quantum physics as being made of photons, these photons are also not electrically charged. Hence X-rays can not be deflected by an electric or magnetic field. Option (d) is correct.
2. Consider a photon of continuous X-ray coming from a Coolidge tube. Its energy comes from
(a) the kinetic energy of the striking electron
(b) the kinetic energy of the free electrons of the target
(c) the kinetic energy of the ions of the target
(d) an atomic transition in the target.
Answer: (a).
Explanation: When a striking electron enters the target, it collides with the atoms of the target before being stopped. During the collision, its kinetic energy is used to increase the temperature of the target or release X-ray photons. This is how continuous X-ray is produced. Option (a) is correct.
3. The energy of a photon of a characteristic X-ray from a Coolidge tube comes from
(a) the kinetic energy of the striking electrons
(b) the kinetic energy of the free electrons of the target
(c) the kinetic energy of the ions of the target
(d) an atomic transition in the target.
Answer: (d).
Explanation: Characteristic X-rays are produced when the striking electrons dislodge inner orbit electrons and the vacancy thus produced in the inner orbit is filled by the transition from outer orbit electrons. The X-ray radiation thus emitted is called characteristic X-ray because its frequency is related to the material of the target. Option (d) is correct.
4. If the potential difference applied to the tube is doubled and the separation between the filament and the target is also doubled the cutoff wavelength
(a) will remain unchanged
(b) will be doubled
(c) will be halved
(d) will become four times the original.
Answer: (c).
Explanation: The cutoff wavelength is given as
𝝀ₘᵢₙ =hc/eV
Where h is the plank's constant, c is the speed of light and V is the potential difference across the Coolidge tube.
When V is doubled,
𝝀'ₘᵢₙ =hc/(e*2V) =½𝝀ₘᵢₙ
So the cutoff wavelength will be halved.
Option (c) is correct.
Note that it is not dependent on the separation between the filament and the target.
5. If the current in the circuit for heating the filament is increased, the cutoff wavelength
(a) will increase
(b) will decrease
(c) will remain unchanged
(d) will change.
Answer: (c).
Explanation: The cutoff wavelength depends upon the potential difference V between the filament and the target, and not on the current in the filament. Since V is unchanged, the cutoff wavelength will remain unchanged. Option (c) is correct.
6. Moseley's law for characteristic X-rays is √𝜈 =a(Z-b). In this
(a) both a and b are independent of the material
(b) a is independent but b depends on the material
(c) b is independent but a depends on the material
(d) both a and b depend on the material.
Answer: (a).
Explanation: Both the constants a and b are independent of the material. The value of a =√(3Rc/4), all constants and the value of b depends upon the orbit from which the transition of an electron is about to take place. Option (a) is correct.
7. Frequencies of Kₐ X-rays of different materials are measured. Which one of the graphs in Figure (44-Q1) may represent the relation between the frequency 𝜈 and the atomic number Z.
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| Figure for Q-7 |
Answer: (d).
Explanation: From Moseley's law,
√𝜈 = a(Z-b)
→𝜈 =a²(Z -b)²
It is an equation of a parabola with some intercept on the Z-axis.
Hence graph d shows the relation between frequency 𝜈 and atomic number Z.
Option (d) is correct.
8. The X-ray beam coming from an X-ray tube
(a) is monochromatic
(b) has all wavelengths smaller than a certain maximum wavelength.
(c) has all wavelengths greater than a certain minimum wavelength
(d) has all wavelengths lying between a minimum and a maximum wavelength.
Answer: (c).
Explanation: There is a certain minimum wavelength called cutoff wavelength for a given potential difference between the filament and the target below which no wavelength is found. All other wavelengths greater than the cutoff wavelength are present in the X-ray beam coming from an X-ray tube. Option (c) is correct.
9. One of the following wavelengths is absent and the rest are present in the X-rays coming from a Coolidge tube. Which one is the absent wavelength?
(a) 25 pm (b) 50 pm
(c) 75 pm (d) 100 pm.
Answer: (a).
Explanation: The wavelengths shorter than the cutoff wavelength are not present in the X-rays coming from a Coolidge tube. If only one wavelength out of the given four wavelengths is absent in the coming X-ray radiation, it must be the shortest one that would be less than the cutoff wavelength. Option (a) is correct.
10. Figure (44-Q2) shows the intensity-wavelength relations of X-rays coming from two different Coolidge tubes. The solid curve represents the relation for tube A in which the potential difference between the target and the filament is VA and the atomic number of the target material is ZA. These quantities are VB and ZB for the other tube. Then
(a) VA > VB, ZA > ZB
(b) VA > VB, ZA < ZB
(c) VA < VB, ZA > ZB
(d) VA < VB, ZA < ZB.
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| Figure for Q-10 |
Answer: (b).
Explanation: Cutoff wavelength is inversely proportional to the potential difference between the target and the filament. The cutoff wavelength shown in the graph is shorter in tube A, hence the potential difference VA > VB.
The Kₐ wavelength in tube B is shorter than the Kₐ wavelength in tube A, which means the Kₐ frequency in tube B is greater than the Kₐ frequency in tube A.
From Moseley's law
Frequency 𝜈 =a²(Z-b)²
Clearly, the Z of the target in tube B is greater than the Z in the target in tube A.
So, ZA < ZB.
So option (b) is correct.
11. 50% of the X-ray coming from a Coolidge tube is able to pass through a 0.1 mm thick aluminum foil. If the potential difference between the target and the filament is increased, the fraction of the X-ray passing through the same foil will be
(a) 0% (b) <50%
(c) 50% (d) >50%.
Answer: (d).
Explanation: The penetrating power of an X-ray photon depends on its energy. When the potential difference between the target and the filament is increased, the cutoff wavelength is decreased which means that more energetic photons are now being produced. These increased energy photons will now fully pass through the given thickness of the aluminum foil. Thus more than 50% of the X-rays will be able to pass through the aluminum foil. Option (d) is correct.
12. 50% of the X-ray coming from a Coolidge tube is able to pass through a 0.1 mm thick aluminum foil. The potential difference between the target and the filament is increased. The thickness of aluminum foil, which will allow 50% of the X-ray to pass through, will be
(a) zero (b) <0.1 mm
(b) 0.1 mm (d) >0.1 mm.
Answer: (d).
Explanation: As we have seen in problem (11) that when the potential difference is increased in the Coolidge tube, more than 50% of the X-rays are able to pass through the 0.1 mm thick aluminum foil. It is due to the fact that more energetic X-rays are now being produced that all pass through the given foil. To limit the fraction of passing through X-rays to 50%, we need to restrict a part of these passing energetic X-rays. For this, we will need a thicker foil. So Option (d) is correct.
13. X-ray from a Coolidge tube is incident on a thin aluminum foil. The intensity of the X-ray transmitted by the foil is found to be Iₒ. The heating current is increased so as to increase the temperature of the filament. The intensity of the X-ray transmitted by the foil will be
(a) zero (b) <Iₒ
(c) Iₒ (d) >Iₒ.
Answer: (d).
Explanation: When the potential difference between the target and the filament is kept the same but the heating current is increased to increase the temperature of the filament, more electrons from the filament are available for being accelerated. The kinetic energy of all electrons reaching the target remains the same. Due to this, the intensity of the produced X-rays increases without increasing the penetrating power. Thus the same fraction of the X-rays will pass through the foil but with increased intensity. So Option (d) is correct.
14. Visible light passing through a circular hole forms a diffraction disc of radius 0.1 mm on a screen. If an X-ray is passed through the same setup, the radius of the diffraction disc will be
(a) zero (b) <0.1 mm
(c) 0.1 mm (d) >0.1 mm.
Answer: (b).
Explanation: The radius of a diffraction disc for a given hole and the distance of the screen are directly proportional to the wavelength of the passing light. Since the wavelength of X-rays is smaller than the wavelength of visible light, the radius of the diffraction disc for X-rays for the same setup will be smaller. Option (b) is correct.
OBJECTIVE-II
1. For harder X-rays,
(a) the wavelength is higher
(b) the intensity is higher
(c) the frequency is higher
(d) the photon energy is higher.
Answer: (c), (d).
Explanation: Harder X-rays mean that X-rays have a comparatively more penetrating power. The more the energy of photons more penetrating power. For higher energy of photons, frequency is also higher. Hence options (c) and (d) are correct.
With higher wavelengths, the energy of photons decreases. So option (a) is not correct.
The increase in intensity does not increase the energy of a photon. Hence option (b) is also incorrect.
2. The cutoff wavelength of X-rays coming from a Coolidge tube depends on the
(a) target material
(b) accelerating voltage
(c) the separation between the target and the filament
(d) the temperature of the filament.
Answer: (b).
Explanation: The cutoff wavelength of X-rays is given as
λₘₐₓ = hc/eV
Here hc/e is constant. Hence the cutoff wavelength of X-rays depends only on the accelerating voltage V. Only (b) is the correct option.
3. Mark the correct options.
(a) an atom with a vacancy has smaller energy than a neutral atom.
(b) K X-ray is emitted when a hole makes a jump from the K shell to some other shell.
(c) The wavelength of K X-ray is smaller than the wavelength of L X-ray of the same material.
(d) The wavelength of Kₐ X-ray is smaller than the wavelength of Kᵦ X-ray of the same material.
Answer: (b), (c).
Explanation: When the parts of atoms are far away we assume their energy is zero. When they are brought together the energy of the atom decreases and expresses as negative. Hence an atom with a vacancy has higher energy than a neutral atom. Option (a) is incorrect.
K X-rays are produced when an electron from another shell jumps to the K shell. Thus the vacancy or hole in the K shell goes to that shell from where the electron has jumped. In other words, the hole has jumped from the K shell to some other shell. Option (b) is correct.
K X-ray has an energy equal to the difference between the K and L shells while L X-ray has an energy equal to the difference between L and M shells. In the same material, the former energy is higher than the latter. Hence option (c) is correct.
Kₐ X-rays are produced when an electron transition takes place from L shell to K shell while Kᵦ X-rays are the result of an electron transition from M shell to K shell. So in the same material, Kₐ X-rays will have lower energy than the Kᵦ X-rays. So Kᵦ X-rays will have a smaller wavelength than the Kₐ X-rays. Option (d) is incorrect.
4. For a given material, the energy and wavelength of characteristic X-rays satisfy,
(a) E(Kₐ) > E(Kᵦ) > E(Kᵧ)
(b) E(Mₐ) > E(Lₐ) > E(Kₐ)
(c) λ(Kₐ) > λ(Kᵦ) > λ(Kᵧ)
(d) λ(Mₐ) > λ(Lₐ) > λ(Kₐ)
Answer: (c), (d).
Explanation: Option (a) is in reverse order, hence incorrect.
As we go higher in shells the energy difference between consecutive shells decreases. Hence the energy given in option (b) is in reverse order, thus incorrect.
For the same material, the energy of Kₐ X-ray is less than the energy of Kᵦ X-ray and it is further less than the energy of Kᵧ X-ray. Hence the wavelength of Kₐ X-ray is greater than the wavelength of Kᵦ X-ray. Also, the wavelength of the Kᵦ X-ray is greater than the Kᵧ X-ray. Hence option (c) is correct.
For the reason explained in the second paragraph above, the wavelength of Mₐ X-ray is greater than the wavelength of Lₐ X-ray, and the wavelength of Lₐ X-ray is greater than the wavelength of Kₐ X-rays. Hence option (d) is correct.
5. The potential difference applied to an X-ray tube is increased. As a result, in the emitted radiation,
(a) the intensity increases
(b) the minimum wavelength increases
(c) the intensity remains unchanged
(d) the minimum wavelength decreases.
Answer: (c), (d).
Explanation: With the increase in the potential difference in the Coolidge tube, the number of striking electrons and hence the emitted photons do not change. Hence the intensity remains unchanged. However, due to increased potential difference, the energy of the striking electrons increases and the cutoff wavelength of emitted X-rays decreases. So options (c) and (d) are correct, not options (a) and (b).
6. When an electron strikes the target in a Coolidge tube, its entire kinetic energy
(a) is converted into a photon
(b) may be converted into a photon
(c) is converted into heat
(d) may be converted into heat.
Answer: (b), (d).
Explanation: When the striking electrons enter the target, they go into many collisions before being stopped. However, all the electrons do not have the same number of collisions and not all in similar ways. A part or whole of the kinetic energy of a colliding electron may increase the vibration of the molecules of the target (Which increases the temperature of the target) and the rest may be converted into photons of X-rays. Hence options (b) and (d) are correct.
7. X-ray incident on a material
(a) exerts a force on it
(b) transfers energy to it
(c) transfers momentum to it
(d) transfers impulse to it.
Answer: All.
Explanation: X-rays are electromagnetic radiation, hence their photons have energy as well as momentum. When X-rays are incident on a material a part or whole of it may be absorbed depending on the nature and thickness of the material. This absorbed part of X-rays transfers their momentum and impulse to the material. Due to this, they exert force on the material. The absorbed X-rays lose their energy to the material that appears as heat. Hence all the options are correct.
8. Consider a photon of continuous X-ray and a photon of characteristic X-ray of the same wavelength. Which of the following is/are different for the two photons?
(a) frequency
(b) energy
(c) penetrating power
(d) method of creation.
Answer: (d).
Explanation: X-rays having the same wavelength have the same frequency, the same energy, and hence the same penetrating power. There will be no difference in the property.
Continuous X-ray is produced due to the loss of kinetic energy of the striking electron while the characteristic X-ray is produced when the striking electron dislodges an inner orbit electron of the atom of the material and this vacancy is filled by any outer orbit electron, thus releasing the photon. So the only difference between both of them is the method of creation. Only option (d) is correct.
EXERCISES
1. Find the energy, frequency, and momentum of an X-ray photon of wavelength 0.10 nm.
ANSWER: Given that wavelength λ =0.10 nm.
So energy,
E =hc/λ
=4.14x10⁻¹⁵*3x10⁸/0.10x10⁻⁹
=124x10² eV
=12.4 keV.
Frequency of the photon
𝜈 =c/λ
=(3x10⁸ m/s)/(0.10x10⁻⁹ m)
=3x10¹⁸ Hz.
The momentum of the X-ray photon
p =E/c
=(12.4x10³ eV)*(1.6x10⁻¹⁹ C)/(3x10⁸ m/s)
=6.61x10⁻²⁴ kg-m/s.
2. Iron emits Kₐ X-ray of energy 6.4 keV and calcium emits Kₐ X-ray of energy 3.69 keV. Calculate the times taken by an iron Kₐ photon and a calcium Kₐ photon to cross through a distance of 3 km.
Answer: Whatever the energy of an X-ray photon, they are all electromagnetic radiations and hence they all travel with the speed of light c. So here both the Kₐ photons will take the same time equal to
(3 km)/(3x10⁸ km/s)
=1.0x10⁻⁸ s
=10 µs.
3. Find the cutoff wavelength for the continuous X-rays coming from the X-ray tube operating at 30 kV.
ANSWER: The potential difference in the X-ray tube,
V =30 kV.
The cutoff wavelength is given as,
λₘᵢₙ = hc/eV
=4.14x10⁻¹⁵*3x10⁸/(30x10³) m
=4.14x10⁻¹¹ m
=41.4 pm.
4. What potential difference should be applied across an X-ray tube to get an X-ray of a wavelength not less than 0.10 nm? What is the maximum energy of a photon of this X-ray in Joule?
ANSWER: Given λₘᵢₙ =0.10 nm.
Since λₘᵢₙ =hc/eV, where V =potential difference
→V =hc/λₘᵢₙ for h in eV-s
=4.14x10⁻¹⁵*3x10⁸/0.10x10⁻⁹ volts
=12.4x10³ volts
=12.4 kV.
The maximum energy of a photon in this X-ray will be equal to the kinetic energy of the accelerated electron in the X-ray tube when all of the kinetic energy is converted into an X-ray photon.
This energy E =eV
→E=1.6x10⁻¹⁹x12.4x10³ J
≈2x10⁻¹⁵ J.
5. The X-ray coming from a Coolidge tube has a cutoff wavelength of 80 pm. Find the kinetic energy of the electrons hitting the target.
ANSWER: Given 𝜆ₘᵢₙ =80 pm.
But 𝜆ₘᵢₙ =hc/eV
Where eV is the kinetic energy of the electrons hitting the target.
→K.E. of electrons, eV =hc/𝜆ₘᵢₙ
=4.14x10⁻¹⁵*3x10⁸/(80x10⁻¹²) eV
=1.55x10⁴ eV
=15.5 keV.
6. If the operating potential in an X-ray tube is increased by 1% by what percentage does the cutoff wavelength decrease?
ANSWER: Cutoff wavelength is given as
λₘᵢₙ =hc/eV
When V is increased by 1%, new V'=1.01V
So now λ'ₘᵢₙ =hc/(1.01eV)
=0.99hc/eV {Approx}
=0.99λₘᵢₙ {Approx)
So the cutoff wavelength decreases by approximately 1%.
7. The distance between the cathode (filament) and the target in an X-ray tube is 1.5 m. If the cutoff wavelength is 30pm, find the electric field between the cathode and the target.
ANSWER: Given λₘᵢₙ =30 pm
Hence from λₘᵢₙ =hc/eV
→30x10⁻¹² =4.14x10⁻¹⁵*3x10⁸/eV
→V =41.4 kV
Distance between the filament and the target, d =1.5 m, hence the field between the two is
E =V/m =41.4x10³/1.5 =27.6 kV/m.
8. The short wavelength limit shifts by 26 pm when the operating voltage in an X-ray tube is increased to 1.5 times the original value. What was the original value of the operating voltage?
ANSWER: Let the original value of the operating voltage =V. New operating voltage,
V' =1.5V.
If the original short wavelength limit =λ pm, then
λ*10⁻¹² =hc/eV
New short wavelength limit λ' =λ -26 pm
So, (λ -26)*10⁻¹² =hc/1.5eV
→hc/eV -26x10⁻¹² =hc/1.5eV
→(hc/eV)*(1 -1/1.5) =26x10⁻¹²
→hc/eV =78x10⁻¹²
→V =4.14x10⁻¹⁵*3x10⁸/78x10⁻¹²
=1.59x10⁴ volts
=15.9 kV.
9. The electron beam in a color TV is accelerated through 32 kV and then strikes the screen. What is the wavelength of the most energetic X-ray photon?
ANSWER: Given V =32 kV.
The wavelength of the most energetic X-ray photon is the cutoff wavelength, that is equal to hc/eV.
Hence the required wavelength,
λₘᵢₙ =hc/eV
=4.14x10⁻¹⁵*3x10⁸/32x10³ m
=0.388x10⁻¹⁰ m
=38.8x10⁻¹² m
=38.8 pm.
10. When 40 kV is applied across an X-ray tube, an X-ray is obtained with a maximum frequency of 9.7x10¹⁸ Hz. Calculate the value of the Plank constant from these data.
ANSWER: Given 𝜈ₘₐₓ =9.7x10¹⁸ Hz.
→𝜆ₘᵢₙ =c/𝜈ₘₐₓ
=3x10⁸/9.7x10¹⁸ m
=3.09x10⁻¹¹ m
Also given V =40 kV
But 𝜆ₘᵢₙ =hc/eV
→h = 𝜆ₘᵢₙ(eV)/c
=3.09x10⁻¹¹*40x10³/3x10⁸ eV-s
=4.12x10⁻¹⁵ eV-s.
11. An X-ray tube operates at 40 kV. Suppose the electron converts 70% of its energy into a photon at each collision. Find the lowest three wavelengths emitted from the tube. Neglect the energy imparted to the atom with which the electron collides.
ANSWER: After each collision, the photon energy will get reduced and a longer wavelength will be produced. Hence for the lowest three wavelengths emitted, we will consider the first three collisions.
The potential difference in the X-ray tube =40 kV.
The energy of the electron just before hitting the target =40 keV.
The energy of the first emitted photon,
E =70% of 40 keV =28 keV.
The wavelength of this emitted radiation
𝜆 =hc/E
=4.14x10⁻¹⁵*3x10⁸/E
=1242x10⁻⁹/E
=1242/E nm
=1242/28x10³ nm
=0.0443 nm
=44.3 pm.
The remaining energy of the electron =40 keV -28 keV =12 keV.
The energy of the photon emitted after the next collision =70% of 12 keV.
=8.4 keV
The wavelength of this emitted radiation,
𝜆 =1242/8.4x10³ nm
=0.148 nm
=148 pm.
Now the remaining energy of the electron =12 keV -8.4 keV =3.6 keV.
The energy of the photon emitted after the next collision =70% of 3.6 keV =2.52 keV.
The wavelength of this emitted radiation
𝜆 =1242/2.52x10³ nm
=0.493 nm
=493 pm.
Further collisions will emit photons with even lower energies. Hence the emitted wavelengths will be longer.
12. The wavelength of Kₐ X-ray of tungsten is 21.3 pm. It takes 11.3 keV to knock out an electron from the L shell of a tungsten atom. What should be the minimum accelerating voltage across an X-ray tube having a tungsten target that allows the production of Kₐ X-ray?
ANSWER: The energy difference between the K shell and the L shell will be equal to the energy of a Kₐ X-ray photon because this photon is emitted when an electron from the L shell fills a vacancy in the K shell of the atom. Here, the energy of a Kₐ X-ray photon,
E =hc/𝜆
=1242/(21.3/1000) eV,
{here wavelength should be in nm}
→E =58.3 keV.
For the production of K alfa X-ray, an electron in the K shell should be knocked out. Since the energy required to knock out an L shell electron is 11.3 keV, the energy required to knock out a K shell electron will be =11.3+58.3 keV =69.6 keV.
So an accelerated electron should have 69.6 keV energy before striking the target for the production of Kₐ X-ray. For an electron to be accelerated to 69.6 keV of energy, the accelerating voltage should be equal to 69.6 V.
13. The Kᵦ X-ray of argon has a wavelength of 0.36 nm. The minimum energy needed to ionize an argon atom is 16 eV. Find the energy needed to knock out an electron from the K shell of an argon atom.
ANSWER: An argon atom has electrons available up to the M shell. Hence the minimum energy to ionize an argon atom will be used to knock out an electron from the outermost M shell.
Kᵦ X-ray is produced when an electron makes a transition from the M shell to a vacancy in the K shell releasing a photon. Hence the energy of a photon of a Kᵦ X-ray will be equal to the difference in energy levels of a K shell and an M shell.
Here given wavelength of Kᵦ X-ray =0.36 nm. Hence the energy of a photon of this X-ray
E =hc/𝜆
=1242/0.36 eV
=3450 eV.
So the energy needed to knock out an electron from the K shell of an argon atom will be
=3450 ev +16 eV
=3466 eV
≈3.47 keV.
14. The Kₐ X-rays of aluminum (Z =13) and Zinc (Z =30) have wavelengths 887 pm and 146 pm respectively. Use Mosley's law √𝜈 =a(Z-b) to find the wavelength of the Kₐ X-ray of iron (Z =26).
ANSWER: 𝜈 =c/𝜆 =3x10⁸/𝜆*10⁻¹² Hz
{for 𝜆 in pm}
→𝜈 =3x10²⁰/𝜆
→√𝜈 =√(3/𝜆) *10¹⁰
Using Moseley's law for the given data for aluminum and zinc, we have
√(3/887) *10¹⁰ =a(13 -b)
→a(13 -b) =5.82x10⁸ ------- (i)
and
√(3/146)x10¹⁰ =a(30 -b)
→a(30-b) =14.33x10⁸ ------- (ii)
Dividing (ii) by (i) we get
(30 -b)/(13 -b) =2.46
→30 -b =13*2.46 -2.46b
→1.46b =31.98 -30 =1.98
→b =1.356
From (ii)
a =14.33x10⁸/(30 -1.356)
=5x10⁷
Hence the wavelength of Kₐ X-ray of Iron, (Z =26) will be given as
√(3/𝜆) *10¹⁰ =a(Z -b)
→√(3/𝜆) *10¹⁰ =5x10⁷*(26-1.356)
→3/𝜆 ={24.644*5/10³}²
=0.0152
→𝜆 =3/0.0152 =197.4 pm.
15. A certain element emits Kₐ X-ray of energy 3.69 keV. Use the data from the previous problem to identify the element.
ANSWER: E =3.69 keV.
Hence the frequency of X-ray radiation,
𝜈 =E/h
=3.69x10³/4.14x10⁻¹⁵ Hz
=8.91x10¹⁷ Hz
From Mosley's law,
√𝜈 =a(Z -b)
→9.44x10⁸ =5x10⁷(Z -1.356)
{The values of constants a and b are taken from the previous problem as directed}
→Z -1.356 =18.87
→Z =18.87 +1.356
≈20
So this is the atomic number of the element calcium.
16. The Kᵦ X-rays from certain elements are given below. Draw the Mosley-type plot of √𝜈 versus Z for Kᵦ radiation.
Element Na P Ca Mn Zn Br
Energy (keV) 0.858 2.14 4.02 6.51 9.57 13.3
ANSWER: Since the energy of a photon is given as
E =h𝜈
→𝜈 =E/h
→√𝜈 =√(E/h)
From the given data, we have the following data for plotting:--
Element Ne P Ca Mn Zn Br
Energy (keV) 0.858 2.14 4.02 6.51 9.57 13.3
√𝜈 (x10⁸) 4.55 7.19 9.85 12.54 15.20 17.92
Z 10 15 20 25 30 35
We plot the graph as follows:--
 |
| Plot for Q-16 |
Z is plotted along the X-axis and √𝜈 along the Y-axis. As we see this graph is a straight line.
17. Use Mosley's law with b =1 to find the frequency of the Kₐ X-ray of La (Z =57) if the frequency of Kₐ X-ray of Cu (Z =29) is known to be 1.88x10¹⁸ Hz.
ANSWER: Given, b =1 and the frequency of Kₐ X-ray 𝜈 =1.88x10¹⁸ Hz for Cu (Z =29).
From Mosley's law,
√𝜈 =a(Z -b)
→√(1.88x10¹⁸) =a(29 -1)
→a =1.37x10⁹/28
=4.9x10⁷
Hence the frequency of Kₐ X-ray of La (Z =57) will be given as
√𝜈 =a(Z -b)
→√𝜈 =4.9x10⁷(57 -1)
→√𝜈 =2.74x10⁹
→𝜈 =7.50x10¹⁸ Hz.
18. Kₐ and Kᵦ X-rays of molybdenum have wavelengths 0.71 Å and 0.63 Å respectively. Find the wavelength of Lₐ X-ray of molybdenum.
ANSWER: From the given data, the wavelengths of Kₐ and Kᵦ X-rays of molybdenum are 0.071 nm and 0.063 nm respectively. Since the Kₐ X-ray results from the transition of an electron from the L shell to the K shell, the energy difference between the K shell and the L shell here is
E(L) -E(K) =hc/λ
=1242/0.071 eV
=17.5 keV
The Kᵦ X-ray results from the transition of an electron from the M shell to the K shell, hence the energy difference between the K shell and M shell here is
E(M) -E(K) =1242/0.063 eV
=19.7 keV
Hence the energy difference between the L shell and the M shell is
E(M) -E(L) =19.7 -17.5 =2.2 keV
We know that the Lₐ X-ray is emitted due to the transition of an electron from the M shell to the L shell. Hence the energy of a photon of the Lₐ X-ray here is
E =E(M) -E(L) =2.2 keV
Hence its wavelength is
𝜆 =1242/(2.2x10³) nm
=0.564 nm
=5.64 Å.
19. The wavelengths of Kₐ and Lₐ X-rays of a material are 21.3 pm and 141 pm respectively. Find the wavelength of Kᵦ X-ray of the material.
ANSWER: The wavelength λ of Kₐ X-ray =21.3 pm =0.0213 nm
Hence E(L) -E(K) =1242/0.0213 eV
=58.3 keV
The wavelength λ of Lₐ X-ray =141 pm =0.141 nm.
Hence E(M) -E(L) =1242/0.141 eV
=8.81 keV
The energy of Kᵦ X-ray =E(M) -E(K)
={E(M) -E(L)} +{E(L) -E(K)}
=8.81 +58.3 =67.11 keV.
So the wavelength of Kᵦ X-ray
λ =1242/(67.11x10³) nm
=0.0185 nm
=18.5 pm.
20. The energy of a silver atom with a vacancy in the K shell is 25.31 keV, the L shell is 3.56 keV, and the M shell is 0.530 keV higher than the energy of the atom with no vacancy. Find the frequency of Kₐ, Kᵦ, and Lₐ X-rays of silver.
ANSWER: From the given data it implies that 25.31 keV of energy is required to separate widely apart an electron from the K shell of the silver atom, 3.56 keV from the L shell, and 0.530 keV from the M shell. Thus the energy difference between the K shell and the L shell is
E(L) -E(K) =25.31 -3.56 =21.75 keV,
Similarly,
E(M) -E(L) = 3.56 -0.53 =3.03 keV, and
E(M) -E(K) =25.31 -0.53 =24.78 keV.
Hence the frequency of Kₐ X-ray
={E(L) -E(K)}/h
=21.75x10³/4.14x10⁻¹⁵ Hz
=5.25x10¹⁸ Hz.
The frequency of Kᵦ X-ray
={E(M) -E(K)}/h
=24.78x10³/4.14x10⁻¹⁵ Hz
=5.98x10¹⁸ Hz.
The frequency of Lₐ X-ray of silver
={E(M) -E(L)}/h
=3.03x10³/4.14x10⁻¹⁵ Hz
=7.32x10¹⁷ Hz.
21. Find the maximum potential difference that may be applied across an X-ray tube with a tungsten target without emitting any characteristic K or L X-ray. The energy levels of the tungsten atom with an electron knocked out are as follows:-
The cell containing vacancy K L M
Energy in keV 69.5 11.3 2.3
ANSWER: For the given condition, the accelerating electron's maximum energy should be just less than the energy required for knocking out an L shell electron. From the given data, the energy required for knocking out an L shell electron from the tungsten atom is 11.3 keV. Hence the striking electron's maximum energy should be less than 11.3 keV which requires a maximum potential difference applied across the given X-ray tube of less than 11.3 kV.
22. The electric current in an X-ray tube (from the target to filament) operating at 40 kV is 10 mA. Assume that an average, of 1% of the total kinetic energy of the electrons hitting the target are converted into X-rays. (a) What is the total power emitted as X-rays and (b) how much heat is produced in the target every second?
ANSWER: The total electric current in the X-ray tube, i =10 mA =0.01 A.
The potential difference is V =40 kV.
Hence the total power being consumed is,
P =i*V =(0.01 A)*(40 kV)
=400 W
(a) Since 1% of the total kinetic energy of the electrons hitting the target is converted into X-rays, hence 1% of the total power consumed will be emitted as X-rays which is equal to
=1% of 400 W =4 W.
(b) Rest of the power is converted into heat i.e. =400 -4 W =396 W.
1 W of power means 1J of energy per second. Hence every second 396 J of energy is converted into heat in the target.
23. Heat at the rate of 200 W is produced in an X-ray tube operating at 20 kV. Find the current in the circuit. Assume that only a small fraction of the kinetic energy of electrons is converted into X-rays.
ANSWER: If the loss of kinetic energy of electrons in converted X-rays is neglected, the total 200 W of power is converted into heat. Operating potential difference V =20kV.
Hence the current in the circuit,
i =Power/P.D.
=(200 W)/(20 kV)
=200/20000 A
=0.01 A
=10 mA.
24. Continuous X-rays are made to strike a tissue paper soaked with polluted water. The incoming X-rays excite the atoms of the sample by knocking out the electrons from the inner shells. Characteristic X-rays are subsequently emitted. The emitted X-rays are analyzed and the intensity is plotted against the wavelength (figure 44-E1). Assuming that only Kₐ intensities are detected, list the elements present in the sample from the plot. Use Mosley's equation
𝜈 =(25x10¹⁴ Hz)(Z -1)².
 |
| The figure for Q-24 |
ANSWER: From the given Mosley's equation,
(25x10¹⁴ Hz)(Z-1)² =𝜈 =c/𝜆
→(Z-1)² =3x10⁸/(25x10¹⁴)𝜆
→Z =1 +(√3/5)x10⁻³/√𝜆)
For 𝜆 =78.9 pm
Z =1 +(√3/5)x10⁻³/√(78.9x10⁻¹²)
=1 +0.039x10³
=1 +39
=40
Hence one of the elements is Zr with Z =40.
For 𝜆 =146 pm,
Z =1 +(√3/5)x10⁻³/√(146x10⁻¹²)
=1 +0.029x10³
=1 +29
=30
Hence one of the elements is Zn with Z =30.
For 𝜆 =158 pm
Z =1 +(√3/5)x10⁻³/√(158x10⁻¹²)
=1 +0.028x10³
=1 +28
=29
So one of the elements is Cu with Z =29.
For 𝜆 =198 pm
Z =1 +(√3/5)x10⁻³/√(198x10⁻¹²)
=1 +0.025x10³
=1 +25
=26
So the remaining element is Fe with Z =26.
Hence the list of elements present in the sample on the basis of the plot is Zr, Zn, Cu, and Fe.
25. A free atom of iron emits Kₐ X-rays of energy 6.4 keV. Calculate the recoil kinetic energy of the atom. Mass of an iron atom =9.3x10⁻²⁶ kg.
ANSWER: Energy of an X-ray photon (given),
E = 6.4 keV.
The linear momentum of this photon,
p =E/c.
After the emission of the photon, the atom will have the same linear momentum but in opposite directions. It is due to the law of conservation of linear momentum. Suppose the recoil speed of the atom is v, then its momentum =mv.
Equating the above two,
mv =p =E/c
Recoiled-kinetic energy of the atom,
K =½mv²
=(mv)²/2m
=p²/2m
=E²/2mc²
=(6.4x10³)²*1.602x10⁻¹⁹/{2*9.3x10⁻²⁶*(3x10⁸)²} eV
{Since the denominator will be in Joule, to convert it to eV the numerator is multiplied by 1.602x10⁻¹⁹}
=3.9x10⁻⁴ eV.
26. The stopping potential in a photoelectric experiment is linearly related to the inverse of the wavelength (1/λ) of the light falling on the cathode. The potential difference applied across an X-ray tube is linearly related to the inverse of the cutoff wavelength (1/λ) of the X-ray emitted. Show that the slopes of the lines in the two cases are equal and find their value.
ANSWER: Stopping potential is given as
Vₒ =(hc/e)(1/λ) -φ/e
Where φ is the work function.
Here the slope of the line is,
m =hc/e
The relation between the potential difference applied across an X-ray tube and the cutoff wavelength λ is given as
λ =hc/eV
→V =(hc/e)(1/λ)
The slope of this line is also the same and
=hc/e
=4.14x10⁻¹⁵*3x10⁸ V-m
=12.42x10⁻⁷ V-m
=1.242x10⁻⁶ V-m.
27. Suppose a monochromatic X-ray beam of wavelength 100 pm is sent through Young's double slit and the interference pattern is observed on a photographic plate placed 40 cm away from the slit. What should be the separation between the slits so that the successive maxima on the screen are separated by a distance of 0.1 mm?
ANSWER: For Young's double-slit interference pattern equation is
w =z𝜆/d
Where 𝜆 is the wavelength of the radiation, d is the separation of the slits, z is the distance of the screen from the slit and w is the spacing of the successive maxima.
Here given that 𝜆 =100 pm =1x10⁻¹⁰ m.
z =40 cm =0.40 m,
w =0.1 mm =1x10⁻⁴ m
and d =?
So, d =z𝜆/w
=0.40*1x10⁻¹⁰/1x10⁻⁴ m
=0.4x10⁻⁶ m
=4x10⁻⁷ m.
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