Bohr's Model and Physics of the Atom
Questions for Short Answer
1. How many wavelengths are emitted by atomic hydrogen in the visible range (380 nm - 780 nm)? In the range of 50 nm to 100 nm?
ANSWER: The wavelengths emitted by atomic hydrogen are given as,
1/λ =R{1/n² -1/m²}, where R is the Rydberg constant.
In the Lyman series, the longest wavelength is for n =1 and m =2,
1/λ =R{1/1² -1/2²} =0.75R
→λ =1/(0.75*1.097x10⁷) m
=1.215x10⁻⁷ m
=121.5 nm
So all other wavelengths in this series are <121.5 nm <380 nm.
For the Balmer series, the longest wavelength is for n =2 and m =3.
1/λ =R{1/2² -1/3²} =5R/36
→λ =36/(5*1.097x10⁷) m
=6.56x10⁻⁷ m =656 nm.
Which is in the visible range.
The next wavelength in this series is for n =2 and m =4.
1/λ =R{1/2² -1/4²} =3R/16
→λ =16/(3*1.097x10⁷) m
=4.86x10⁻⁷ m =486 nm.
This is also in the visible range.
The next wavelength in this series is for n =2 and m =5.
So 1/λ =R{1/2² -1/5²} =21R/100
→λ =100/(21x1.097x10⁷) m
=4.34x10⁻⁷ m =434 nm.
It is also in the visible range.
Next for n =2 and m =6,
1/λ =R{1/2² -1/6²} =8R/36 =2R/9
→λ =9/(2*1.097x10⁷) m
=4.10x10⁻⁷ m =410 nm.
It is also in the visible range.
Next for n =2 and m =7,
we get λ =(4x49)/45R
=397 nm.
It is also in the visible range.
For n =2 and m =8,
λ =64/15R =389 nm. It is more than 380 nm, so also in the visible range.
Next for n =2 and m =9,
λ = (81*4)/77R =384 nm, also in the visible range because it is more than 380 nm.
For n =2 and m =10,
λ =100/24R =379.6 nm which is just less than 380 nm. So not in the range.
Thus in the Balmer series, there are eight emitted wavelengths between 380 nm to 780 nm.
Let us check in the Paschen series. The shortest wavelength in this series is for n =3 and m =∞.
So λ =9/R =820 nm. Which is more than 780 nm and not in the visible range. All other wavelengths in this series will be longer than it and not in the visible range.
Thus a total of eight wavelengths are emitted by atomic hydrogen in the given visible range.
As we have seen in the beginning, the longest wavelength in the Lyman series is 121.5 nm, the wavelengths between 50 nm to 100 nm will be found in this series only. Here n =1. Let us calculate for m =3.
1/λ =R{1/1² -1/3²} =8R/9
→λ =9/{8*1.097x10⁷} m
=1.025x10⁻⁷ m =102.5 nm
For m =4, λ =16/15R =97.2 nm.
For m =5, λ =25/24R =95 nm.
For m =6, λ =36/35R =93.7 nm.
and so on till m =∞, where λ =1/R
→λ =91.1 nm.
Thus we see that for m >4 to infinite there is an infinite number of closely spaced wavelengths between 50 nm to 100 nm.
2. The first excited energy of a He⁺ ion is the same as the ground state energy of hydrogen. Is it always true that one of the energies of any hydrogen-like ion will be the same as the ground state energy of a hydrogen atom?
ANSWER: The ground state energy of hydrogen is given as,
E = -(13.6 eV)Z²/n² =-13.6 eV
{Since Z =1, n =1}
For the first excited state energy of He⁺ ion, (Z =2, n =2),
E' =-13.6 eV
E =E'.
We see that if only Z =n, (n-1)th excited state of the hydrogen-like ion will be equal to the ground state energy of the hydrogen ion. So it is not always true.
3. Which wavelengths will be emitted by a sample of atomic hydrogen gas (in the ground state) if electrons of energy 12.2 eV collide with the atoms of the gas?
ANSWER: In collisions, the hydrogen atom may absorb a part of the energy of the moving electrons to reach higher excited states. If we assume that after the collision the maximum energy of the colliding electron 12.2 eV is absorbed by the hydrogen atom then the total energy of the hydrogen atom now is,
E =-13.6 eV +12.2 eV =-1.4 eV.
Suppose with this energy the electron of the hydrogen atom jumps to the nth orbit. But the energy of the excited hydrogen atom with the electron in the nth orbit is given as,
E =-13.6/n² eV.
Equating,
-1.4 =-13.6/n²
→n² =13.6/1.4 =9.71
→n =3.1
Since n can take only integer values, the colliding electrons can transfer only that much energy which is needed to take the atom to 1st or 2nd excited state, i.e. n =2 or 3.
When the electron returns back to the ground state i.e. from n =3 to n =1, and n =2 to n =1, the two wavelengths of radiation emitted will be in the wavelength range of the Lyman series.
Wavelength of radiation for n =1, m =3;
1/λ =R{1/1² -1/3²}
=8R/9
→λ =9/8R
=9/(8*1.097x10⁷) m
=1.025x10⁻⁷ m
=102.5 nm.
Wavelength of radiation for n =1, m =2;
1/λ =R{1/1² -1/2²}
→λ =4/3R
=4/(3*1.097x10⁷) m
=1.215x10⁻⁷ m
=121.5 nm.
4. When white radiation is passed through a sample of hydrogen gas at room temperature, absorption lines are observed in the Lyman series only. Explain?
ANSWER: White radiation is X-ray radiation having photon energies of more than 41.4 eV. Hydrogen gas at room temperature is in the ground state i.e. n =1. When the white radiation is passed through the hydrogen gas at room temperature, the hydrogen atoms may absorb that much energy from the colliding photons which are required to take its electrons to higher energy levels, i.e. from n =1 to n =2, 3, 4, ....
Hence in the absorption spectrum, wavelengths corresponding to these energy absorptions (λ =hc/E) are missing and such absorption lines are observed. These are the same wavelengths that are found in the emission phase when electrons jump back to the ground state (n =1) from a higher energy level. These wavelengths are named as Lyman series. Other series of wavelengths are not observed in the absorption spectra because energies are not being absorbed from excited states of hydrogen atoms i.e. from n =2, 3, ....
5. The Balmer series was observed and analyzed before the other series. Can you suggest a reason for such an order?
ANSWER: The Balmer series corresponds to radiations due to the jumping back of electrons from higher energy levels to n =2. These wavelengths range from 656.3 nm to 365 nm. Since most parts of the Balmer series fall in the visible range of 380 nm to 780 nm, it was natural to be observed and analyzed before other series.
6. What will be the energy corresponding to the first excited state of hydrogen if the potential energy of the atom is taken to be 10 eV when the electron is widely separated from the proton? Can we still write Eₙ =E₁/n²? rₙ =aₒn²?
ANSWER: Normally when the electron is widely separated from the proton, the potential energy of the atom is assumed to be zero. In this case, since the potential energy is already 10 eV, the energy of the first excited state of the hydrogen atom will be,
=-10 -13.6/2² eV
=-10-3.4 eV
=-13.4 eV.
We can still write Eₙ =E₁/n² and rₙ =aₒn² because the effects of the reference point for potential energy have already been taken in calculating E₁ and aₒ.
7. The difference in the frequencies of the series limit of the Lyman series and the Balmer series is equal to the frequency of the first line of the Lyman series. Explain.
ANSWER: The series limit is the highest frequency of that series. So for the series limit, the energy of the photons is maximum in that series. Electrons will release maximum energy if they jump down to the highest number of energy levels.
So the energy limit frequency in the Lyman series will be found when electrons jump down from n =∞ to n=1. And in the Balmer series from n =∞ to n =2.
So energy limit of the Lyman series,
f =k{1/n² -1/m²}, where k is a constant.
=k{1/1² -1/∞²}
=k.
And the energy limit of the Balmer series,
f' =k{1/2² -1/∞²}
=k/4
Now, the first line of the Lyman series is the shortest frequency of the series corresponding to the jumping down of electrons from n =2 to n =1. This frequency is,
f" =k{1/1² -1/2²}
=3k/4.
We see that the difference in the series limits of these two series is
=k -k/4
=3k/4
=f".
8. The numerical value of ionization energy in eV equals the ionization potential in volts. Does the equality hold if these quantities are measured in some other units?
ANSWER: The potential difference through which an electron should be accelerated to acquire that much energy that is required to completely detach it from the nucleus is called ionization potential. So the unit of ionization potential is Volts. Now the energy required to completely detach an electron from the nucleus is the ionization energy. When we express this ionization energy in eV, then by definition it is the energy required to accelerate an electron through a potential difference of V volts. So, numerically ionization energy and ionization potential are equal.
It will not be the case if the ionization energy is expressed in other units.
9. We have stimulated emissions and spontaneous emissions. Do we also have stimulated absorption and spontaneous absorption?
ANSWER: When an atom in a higher energy state is left there, it will eventually come down to a lower state by emitting a photon of energy that is equal to the difference in energy levels. This emission is called spontaneous emission. But when an atom emits a photon due to its interaction with a photon incident on it, the process is called stimulated emission. The emitted photon has exactly the same energy, phase, and direction as the incident photon.
When an atom in a lower energy state is incident upon by a photon having energy equal to the difference of energy levels, it may absorb the photon and jump to a higher energy state. This process is called stimulated absorption. However, there can not be spontaneous absorption because an atom can not jump to a higher energy state itself without absorbing a photon.
10. An atom is in its excited state. Does the probability of its coming to the ground state depend on whether the radiation is already present or not? If yes does it also depend on the wavelength of the radiation present?
ANSWER: An atom in its excited state does not remain so for long if it is left to itself. It eventually comes down to the ground state after the "lifetime of that state" by emitting a photon. But when radiation is present, the excited atom may interact with the photon and come down to the ground state by emitting a similar photon. It is called stimulated emission. So the probability of the atom coming down to the ground state depends upon the radiation present.
The energy of a photon depends upon the wavelength. In the stimulated emission the energy of the incident photon should be equal to the difference in the energy states because a similar photon is emitted after the interaction. So the coming down to the ground state in the presence of radiation also depends upon the wavelength of the radiation.
OBJECTIVE-I
1. The minimum orbital angular momentum of the electron in a hydrogen atom is
(a) h
(b) h/2
(c) h/2π
(d) h/λ
ANSWER: (c).
EXPLANATION: According to Bohr's Postulates, the orbital angular momentum l of the electron about the nucleus is an integral multiple of Planck's constant h divided by 2π. i.e.
l =n*(h/2π)
The minimum value of l is for n =1 and it is equal to h/2π. Option (c) is correct.
2. Three photons coming from the excited atomic-hydrogen sample are picked up. Their energies are 12.1 eV, 10.2 eV, and 1.9 eV. These photons must come from
(a) a single atom
(b) two atoms
(c) three atoms
(d) either two atoms or three atoms.
ANSWER: (d).
EXPLANATION: When an electron jumps to a lower orbit of an excited hydrogen atom, the energy of the photon released is given as,
E = 13.6(1/n² -1/n'²).
For n =1 and n' =3,
E =13.6(1/1² -1/3²) =12.1 eV
For n =1 and n' =2,
E =13.6(1 -1/2²) =10.2 eV
For n =2 and n' =3,
E =13.6(1/2² -1/3²) =1.9 eV
The three picked-up electrons have these values of energy. So one electron has jumped down from the third energy level to the first in an atom. The other two energies are for an electron jumping from the third to the second energy level and from the second to the first energy level. This might have happened in one atom or in two atoms separately. So the electrons are either from two atoms or three atoms. Option (d) is correct.
3. Suppose the electron in a hydrogen atom makes a transition from n =3 to n =2 in 10⁻⁸ s. The order of the torque acting on the electron in this period, using the relation between torque and angular momentum as discussed in the chapter on rotational mechanics is
(a) 10⁻³⁴ N-m
(b) 10⁻²⁴ N-m
(c) 10⁻⁴² N-m
(d) 10⁻⁸ N-m.
ANSWER: (b).
EXPLANATION: The orbital angular momentum of an electron is nh/2π. Change of angular momentum in transition from n =3 to n =2 is
L =3h/2π -2h/2π
=h/2π
Time in this transition t =10⁻⁸ s.
The rate of change of angular momentum is equal to the torque.
Hence torque =L/t
=h/2πt
=6.63x10⁻³⁴/(2*3.14*10⁻⁸) N-m
=1.05x10⁻²⁶ N-m
=(1.05/100)x10⁻²⁴ N-m
So it is of the order of 10⁻²⁴ N-m. option (b) is correct.
4. In which of the following transitions will the wavelength be minimum?
(a) n =5 to n =4
(b) n =4 to n =3
(c) n =3 to n =2
(d) n =2 to n =1.
ANSWER: (d).
EXPLANATION: Wavelength will be minimum for the highest energy released.
Since the energy released is proportional to (1/n² -1/m²)
i.e. =(m²-n²)/m²n²
=(m+n)(m-n)/m²n²
In the given options, m-n =1 for each case. So the energy is proportional to (m+n)/m²n². Clearly, the energy will be maximum for the lowest values of m and n, that is m =2 and n =1. Option (d) is correct.
5. In which of the following systems will the radius of the first orbit (n =1) be minimum?
(a) hydrogen atom
(b) deuterium atom
(c) singly ionized helium
(d) doubly ionized lithium.
ANSWER: (d).
EXPLANATION: The radius of an orbit in a hydrogen-like atom is given by
rₙ =n²aₒ/Z
where Z is the number of protons in the nucleus.
For the first orbit, n =1, So
r₁ =aₒ/Z
Clearly, it will be the minimum for the maximum Z. Here Z is the maximum for lithium which is 3. So the minimum radius of the first orbit will be for doubly ionized lithium. Option (d) is correct.
6. In which of the following systems will the wavelength corresponding to n =2 to n =1 be minimum?
(a) hydrogen atom
(b) deuterium atom
(c) singly ionized helium
(d) doubly ionized lithium.
ANSWER: (d).
EXPLANATION: The wavelength corresponding to the transition from n' to n orbit is given as
1/𝜆 =RZ²{1/n² -1/n'²}
where R is Rydberg's constant and Z is the atomic number. Here n =1 and n' =2, so
1/𝜆 =3RZ²/4
→𝜆 =4/(3RZ²)
So the wavelength will be minimum for maximum Z which is for doubly ionized lithium. Option (d) is correct.
7. Which of the following curves may represent the speed of the electron in a hydrogen atom as a function of the principal quantum number n? 
The figure for Q-7
ANSWER: (c).
EXPLANATION: The speed of an electron in a hydrogen atom is given as,
v =Ze²/(2εₒhn)
Where Z is the number of protons in the nucleus and n is the principal quantum number.
Clearly, the speed is inversely proportional to the principal quantum number n. So the curve v versus n will be a rectangular hyperbola that resembles (c) in the picture. So option (c) is correct.
8. As one considers orbits with higher values of n in a hydrogen atom, the electric potential energy of the atom
(a) decreases
(b) increases
(c) remains the same
(d) does not increase.
ANSWER: (b).
EXPLANATION: The electric potential energy of a hydrogen-like atom is given as,
V =-mZ²e⁴/(4εₒ²h²n²)
=-2*13.6/n² eV
{For a hydrogen atom where Z =1}
So with the higher value of n, the electric potential energy of the atom increases. Option (b) is correct.
9. The energy of an atom (or ion) in the ground state is -54.4 eV. It may be
(a) hydrogen
(b) deuterium
(c) He⁺
(d) Li⁺⁺.
ANSWER: (c).
EXPLANATION: The total energy of a hydrogen-like atom is,
E =-13.6Z²/n²
Given that, E =-54.4 eV, for the ground state, n =1. So
-54.4 = -13.6Z²/1²
→Z² =4.06
→Z ≈2
So the atom/(ion) has two numbers of protons in the nucleus. Hence it must be He⁺ ion. Option (c) is correct.
10. The radius of the shortest orbit in a one-electron system is 18 pm. It may be
(a) hydrogen
(b) deuterium
(c) He⁺
(d) Li⁺⁺.
ANSWER: (d).
EXPLANATION: For a one-electron system ion, the radius of an orbit is given as,
rₙ =n²aₒ/Z
Where aₒ is the radius of the shortest orbit in a hydrogen atom and its value is 53 pm.
Given for the atom, r₁ =18 pm, n =1, hence,
18 =1²*53/Z
→Z =53/18 ≈ 3.
So it is a Li⁺⁺ ion. Option (d) is correct.
11. A hydrogen atom in the ground state absorbs 10.2 eV of energy. The orbital angular momentum of the electron is increased by
(a) 1.05x10⁻³⁴ J-s
(b) 2.11x10⁻³⁴ J-s
(c) 3.16x10⁻³⁴ J-s
(d) 4.22x10⁻³⁴ J-s.
ANSWER: (a).
EXPLANATION: In the ground state energy of a hydrogen atom,
E =-13.6/n² eV =-13.6 eV
In the next excited state (for n =2) the energy,
E' =-13.6/2² eV =-3.4 eV
The difference in the energies of the two states =E'-E
→ =-3.4 -(-13.6) eV =10.2 eV.
So after absorbing 10.2 eV of energy, the hydrogen atom in the ground state is excited to the next energy state and the electron jumps to the 2nd orbit i.e. n =2. The orbital angular momentum of an electron corresponding to the orbits are
=nh/2π
Since the electron jumps from n =1 to n =2, the orbital angular momentum increases by {2h/2π -h/2π}
=h/2π
=6.63x10⁻³⁴/2π J-s
=1.05x10⁻³⁴ J-s
Option (a) is correct.
12. Which of the following parameters are the same for all hydrogen-like atoms and ions in their ground states?
(a) radius and orbit
(b) speed of the electron
(c) the energy of the atom
(d) orbital angular momentum of the electron.
ANSWER: (d).
EXPLANATION: The radius of an orbit in a hydrogen-like atom in the ground state is
rₙ =n²aₒ/Z
→r₁ =aₒ/Z.
It depends on the number of protons in the nucleus. Option (a) is not correct.
Speed of an electron in the ground state,
v =Ze²/(2εₒh)
it also depends on Z, so option (b) is not correct.
The energy of the atom in the ground state is =-13.6Z² eV
Again it depends on Z, so option (c) is not correct.
The orbital angular momentum in the ground state is =nh/2π =h/2π. It is a constant. So Option (d) is correct.
13. In a laser tube, all the photons
(a) have the same wavelength
(b) have same energy
(c) move in the same direction
(d) move with the same speed.
ANSWER: (d).
EXPLANATION: The photons in a LASER tube have a very small difference in wavelengths that are called spread Δλ. So their wavelengths and energies are not exactly the same. Options (a) and (b) are not correct.
The directions of the photons are almost parallel but not exactly parallel. So option (c) is also not correct.
Since the photons in the LASER tube are electromagnetic radiation, they all travel with the speed of light. Hence option (d) is correct.
OBJECTIVE-II
1. In a laboratory experiment on emission from atomic hydrogen in a discharge tube, only a small number of lines are observed whereas a large number of lines are present in the hydrogen spectrum of a star. This is because in a laboratory
(a) the amount of hydrogen taken is much smaller than that present in the star
(b) the temperature of hydrogen is much smaller than that of the star
(c) the pressure of hydrogen is much smaller than that of the star
(d) the gravitational pull is much smaller than that of the star.
ANSWER: (b).
EXPLANATION: The temperature of a star is much higher than that of the laboratory experiment. Hence the electrons get much higher energies to jump to the orbits of much higher energy levels. So when they jump down to lower orbits, they have many energy levels available. Thus radiations have a large range of wavelengths, so a large number of lines are present in the hydrogen spectrum. Option (b) is correct.
2. An electron with kinetic energy 5 eV is incident on a hydrogen atom in its ground state. The collision
(a) must be elastic
(b) may be partially elastic
(c) must be completely inelastic
(d) maybe completely inelastic.
ANSWER: (a).
EXPLANATION: The energy of a hydrogen atom in its ground state is -13.6 eV. The first excited state has an energy of -3.4 eV. So if the collision is inelastic, it needs to absorb at least energy equal to -3.4 -(-13.6) =10.2 eV. Thus in this case the colliding electron must have energy equal to 10.2 eV for completely inelastic collision or more than 10.2 eV for partially inelastic collision. In this case, the incident electron's energy is only 5 eV which can not be absorbed by the atom. So the collision will be elastic.
Option (a) is correct.
3. Which of the following products in a hydrogen atom are independent of the Principal quantum number n? The symbols have their usual meanings.
(a) vn
(b) Er
(c) En
(d) vr.
ANSWER: (a), (b).
EXPLANATION: The relations among E, n, r, and v for a hydrogen-like atom are as follows.
E =-mZ²e⁴/(8εₒ²h²n²) ----- (i)
r =εₒh²n²/(πmZe²) --------- (ii)
v =Ze²/(2εₒhn) ------------- (iii)
From (iii), vn =Ze²/(2εₒh). It is independent of n. Option (a) is correct.
From (i) and (ii),
Er =-Ze²/(8πεₒ). It is also independent of n. So Option (b) is also correct.
From (i), En =-mZ²e⁴/(8εₒh²n). Clearly, this product depends on n. Option (c) is not correct.
From (ii) and (iii),
vr =hn/(2πm). So this product also depends on n. Option (d) is not correct.
4. Let Aₙ be the area enclosed by the nth orbit in a hydrogen atom. The graph of ln(Aₙ/A₁) against ln(n)
(a) will pass through the origin
(b) will be a straight line with slope 4
(c) will be a monotonically increasing nonlinear curve
(d) will be a circle.
ANSWER: (a), (b).
EXPLANATION: The area of the nth orbit is,
Aₙ =πrₙ², and the area of the first orbit
A₁ =πr₁².
But rₙ=n²aₒ.
So, r₁ =aₒ.
Thus, Aₙ/A₁ =n⁴
→ln (Aₙ/A₁) =ln(n⁴) =4 ln(n)
It is in the form of y =4x. So the graph of ln (Aₙ/A₁) vs ln (n) will be a straight line that will pass through the origin with slope 4.
Options (a) and (b) are only correct.
5. The ionization energy of a hydrogen-like ion A is greater than that of another hydrogen-like ion B. Let r, u, E, and L represent the radius of the orbit, speed of the electron, energy of the atom, and orbital angular momentum of the electron respectively. In ground state
(a) rA > rB
(b) uA > uB
(c) EA > EB
(d) LA > LB.
ANSWER: (b).
EXPLANATION: The ionization energy of a hydrogen-like ion with atomic number Z is given as,
V =(13.6 eV)*Z².
Hence the atomic number of A is greater than B. The radius of the orbit is inversely proportional to Z, so rA < rB.
Option (a) is incorrect.
The speed of the electron is directly proportional to Z. Hence uA > uB.
Option (b) is correct.
The energy of the ion is
E =-mZ²e⁴/(8εₒ²h²n²)
Due to the negative sign, the Energy of A is less than B.
Option (c) is incorrect.
The orbital angular momentum does not depend on Z, it will be the same for ions in the ground state.
Option (d) is also incorrect.
6. When a photon stimulates the emission of another photon, the two photons have
(a) same energy
(b) same direction
(c) same phase
(d) same wavelength.
ANSWER: All.
EXPLANATION: When an atom emits a photon due to its interaction with a photon incident on it, the process is called stimulated emission. The emitted photon has exactly the same energy (hence the same wavelength), phase, and direction as the incident photon.
So, all the options are correct.
Exercises
1. The Bohr radius is given by aₒ =εₒh²/πme². Verify that the RHS has dimensions of length.
ANSWER: Dimensions of the entities on RHS are as follows:-
Dimensions of εₒ =[M⁻¹L⁻³T⁴I²]
Dimensions of h =[ML²T⁻¹]
π is dimensionless
Dimension p is m =[M]
Dimensions of e =[IT]
Hence the dimensions of RHS are,
[M⁻¹L⁻³T⁴I²][ML²T⁻¹]²/[M][I²T²]
=[MLT²I²]/[MI²T²]
=[L] =Dimension of length.
2. Find the wavelength of the radiation emitted by hydrogen in the transitions (a) n =3 to n =2, (b) n =5 to n =4 and (c) n =10 to n =9.
ANSWER: (a) The wavelength is given as
1/𝜆 =R{1/n² -1/m²}
Here n =2 and m =3,
So 1/𝜆 =R{1/4 -1/9} =5R/36
=5*1.097x10⁷/36
→𝜆 =656x10⁻⁹ m =656 nm.
(b) Here, n =4, m =5, hence
1/𝜆 =R{1/16 -1/25} =9R/(16*25)
→𝜆 =400/9*1.097x10⁷ m
=4050x10⁻⁹ m
=4050 nm.
(c) Here n =9 and m =10.
So 1/𝜆 =R{1/81 -1/100}
=19R/8100
→𝜆 =8100/19*1.097x10⁷ m
=38862x10⁻⁹ m
=38862 nm.
3. Calculate the smallest wavelength of radiation that may be emitted by (a) hydrogen, (b) He⁺, and (c) Li⁺⁺.
ANSWER: The smallest wavelength will be when the energy released by the down-jumping electron is maximum. The energy dissipated will be maximum when the electron jumps from n = infinity to n =1, i.e. to the ground state. For hydrogen or hydrogen-like atoms, the wavelength of the emitted radiation is given as
1/𝜆 =RZ²{1/n² -1/m²}
where R is Rydberg's constant and Z is the atomic number.
(a) For hydrogen, Z =1, n =1 and m =∞. Hence the smallest wavelength is given as
1/𝜆 =R{1/1² -1/∞²} =R
→𝜆 =1/R
=1/1.097x10⁷ m
=91x10⁻⁹ m
=91 nm.
(b) For He⁺, Z =2. Hence the smallest wavelength is given as
1/𝜆 =R*2²{1/1² -1/∞²} =4R
→𝜆 =1/4R
=1/4*1.097x10⁷ m
=23x10⁻⁹ m
=23 nm.
(c) For Li⁺⁺ Z =3. Hence the smallest wavelength is given as,
1/𝜆 =R*3²{1/1² -1/∞²}
=9R
→𝜆 =1/9*1.097x10⁷ m
=10x10⁻⁹ m
=10 nm.
4. Evaluate the Rydberg constant by putting the values of the fundamental constants in its expression.
ANSWER: Expression for Rydberg constant R is given as
R =me⁴/8εₒ²h³c
Value of εₒ =8.85x10⁻¹² C²-m/N
mass of an electron, m =9.1x10⁻³¹ kg
Charge on an electron, e =1.6x10⁻¹⁹ C
Planks constant h =6.63x10⁻³⁴ J-s
Speed of light c =3x10⁸ m/s
Hence,
R =9.1x10⁻³¹*(1.6x10⁻¹⁹)⁴/8*(8.85x10⁻¹²)²*(6.63x10⁻³⁴)³*3x10⁸ m⁻¹
≈1.09x10⁷ m⁻¹.
5. Find the binding energy of a hydrogen atom in the state n =2.
ANSWER: The binding energy of an atom in a given state is the energy needed to separate its constituents over large distances. Since the energy of a hydrogen atom for a state n is
E =-13.6/n² eV, it will require energy equal to 13.6/n² eV to separate the electron from the nucleus over a large distance. Hence the binding energy for the hydrogen atom in the state n =2 is
E' =13.6/2² eV =3·4 eV.
6. Find the radius and energy of a He⁺ ion in the states (a) n =1, (b) n =4, and (c) n =10.
ANSWER: The radius and energy of a hydrogen-like ion are given as,
rₙ =n²aₒ/Z, and Eₙ =-13.6Z²/n²
For He⁺ ion, Z =2,
Hence, rₙ =½n²aₒ, and Eₙ =-54.4/n² eV.
(a) For n =1, radius
r₁ =½*1²*aₒ
=½*53x10⁻¹² m
=0.265x10⁻¹⁰ m
=0.265 Å.
And the energy,
E =-54.4/1² eV
=-54.4 eV.
(b) For n =4.
Radius r =½*4²*aₒ
→r =8*53x10⁻¹² m
=4.24x10⁻¹⁰ m
=4.24 Å.
Energy, E =-54.4/4² eV
=-3.4 eV.
(c) For n =10,
Radius r =½*10²*53x10⁻¹² m
=26.5x10⁻¹⁰ m
=26.5 Å.
Energy E =-54.4/10² eV
=-0.544 eV.
7. A hydrogen atom emits ultraviolet radiation of wavelength 102.5 nm. What are the quantum numbers of the states involved in the transition?
ANSWER: Since the emitted wavelength is in the ultraviolet region of the spectrum, it falls under the Lyman series. Here the electrons jump down to the ground state, n =1. Given that λ =102.5 nm.
Rydberg's constant R =1.097x10⁷ m⁻¹
Let us assume that the transition of the electron is from quantum number n =m to n =1. Hence,
1/λ =R{1/1² -1/m²}
→1/102.5x10⁻⁹ =1.097x10⁷(1-1/m²)
→1 -1/m² =0.889
→1/m² =1-0.889 =0.111
→ m² =9
→m =3.
So the quantum numbers of the states involved in the transition are 1 and 3.
8. (a) Find the first excitation potential of He⁺ ion. (b) Find the ionization potential of Li⁺⁺ ion.
ANSWER: (a) The excitation energy for the He⁺ ion for the first excited state is
E =-13.6Z²{1/1² -1/2²} eV
=-10.2*Z² eV
=-10.2*2² eV {Here Z =2}
=-40.8 eV.
Hence the first ionization potential of the He⁺ ion is 40.8 V.
(b) For Li⁺⁺ ion, Z =3.
Ionization energy here is the energy needed to widely separate the electron from the nucleus which is equal to
=-13.6*Z²
=-13.6*3² eV
=-122.4 eV.
Hence the ionization potential is 122.4 V.
9. A group of hydrogen atoms is prepared in n =4 states. List the wavelengths that are emitted as the atoms make transitions and return to n =2 states.
ANSWER: Three types of wavelengths will be emitted for the transitions from
(a) n =4 to n =2
(b) n =4 to n =3
then (c) n =3 to n =2.
Wavelength for transition from n =4 to n =2 will be given as,
1/𝜆 =R{1/2² -1/4²}
=1.097x10⁷*3/16
=2.057x10⁶
→𝜆 =487x10⁻⁹ m
=487 nm.
The wavelength for the transition from n =4 to n =3 will be given as
1/𝜆 =R{1/3² -1/4²}
=1.097x10⁷*7/(9*16) m
=5.33x10⁵ m
→𝜆 =1876x10⁻⁹ m
=1876 nm.
The wavelength for the transition from n =3 to n =2 will be given as
1/𝜆 =R{1/2² -1/3²}
=1.097x10⁷*5/36
=1.524x10⁶
→𝜆 =656x10⁻⁹ m
=656 nm.
10. A positive ion having just one electron ejects it if a photon of wavelength 228 Å or less is absorbed by it. Identify the ion.
ANSWER: Since the ion ejects the electron after absorbing the photon, the ionization energy of the ion should be equal to the energy of the photon.
𝜆 =228 Å =228x10⁻¹⁰ m
The energy of the photon,
E = hc/𝜆
=6.63x10⁻³⁴*3x10⁸/228x10⁻¹⁰ J
=8.72x10⁻¹⁸ J.
The ionization energy of an ion is given as
E' =13.6Z² eV,
{where Z is the atomic number of the ion.}
→E' =13.6x1.6x10⁻¹⁹Z² J.
Equating both the energies to solve for Z we get
Z² =8.72x10⁻¹⁸/(13.6*1.6x10⁻¹⁹)
→Z² =4
→Z =2.
Hence the ion, in this case, is He⁺.
11. Find the maximum Coulomb force that can act on the electron due to the nucleus in a hydrogen atom.
ANSWER: Coulomb force will be maximum when the electron is in the nearest possible orbit and it is in the ground state for which n =1.
The radius of the orbit is given as,
r =εₒh²n²/(πmZe²)
For the hydrogen atom Z =1 and here n =1,
So, r =εₒh²/(πme²)
=8.85x10⁻¹²*(6.63x10⁻³⁴)²/{π9.1x10⁻³¹*(1.6x10⁻¹⁹)²} m
=5.31x10⁻¹¹ m
There is only a proton in the nucleus of the hydrogen, the charge on both the electron and the proton is e with opposite in nature, hence the Coulomb force on the electron is,
F =(1/4πεₒr²)e²
=9x10⁹*(1.6x10⁻¹⁹)²/(5.31x10⁻¹¹)² N
=8.2x10⁻⁸ N.
12. A hydrogen atom in a state having a binding energy of 0.85 eV makes the transition to a state with an excitation energy of 10.2 eV. (a) Identify the quantum numbers n of the upper and lower energy states involved in the transition. (b) Find the wavelength of the emitted radiation.
ANSWER: (a) The energy in a state of a hydrogen atom is numerically equal to the binding energy in that state. The energy in a state n is given as,
Eₙ =E₁/n², where E₁ is the energy in the ground state that is equal to -13.6 eV. Hence
Eₙ =-13.6/n².
Given binding energy =0.85 eV, the energy in this state, Eₙ =-0.85 eV.
Hence,
n² =-13.6/Eₙ =-13.6/-0.85 =16
→n =4.
Given that for the final state, excitation energy =10.2 eV. Let the quantum number of this state =m. So, the energy of this state =-13.6 +10.2 eV =-3.4 eV.
Thus, m² =-13.6/-3.4 =4
→m =2.
So in this transition, the upper energy state is 4 and the lower energy state is 2.
(b) The wavelength 𝜆 of the emitted radiation is given as,
1/𝜆 =R{1/n² -1/m²}
Where m > n, So here n =2 and m =4
→1/𝜆 =1.097x10⁷{1/2² -1/4²}
→1/𝜆 =(3/16)*1.097x10⁷ m
→𝜆 =486x10⁻⁹ m =486 nm.
13. Whenever a photon is emitted by hydrogen in the Balmer series, it is followed by another photon in the Lyman series. What wavelength does this later photon correspond to?
ANSWER: Since the photon is emitted in the Balmer series, the electron has jumped to the second energy level n =2. The following photon is emitted in the Lyman series, which means the electron here has jumped from n =2 to n =1. So the wavelength corresponding to the later emitted is given as
1/𝜆 =R{1/1² -1/2²} =3R/4
→𝜆 =4/3R
=4/{3*1.097x10⁷} m
=1.22x10⁻⁷ m
=122x10⁻⁹ m
=122 nm.
14. A hydrogen atom in state n =6 makes two successive transitions and reaches the ground state. In the first transition, a photon of 1.13 eV is emitted. (a) Find the energy of the photon emitted in the second transition. (b) what is the value of n in the intermediate state?
ANSWER: The energy of the hydrogen atom in state n =6 is
E₆ =-13.6/n² =-13.6/6² eV
=-0.38 eV.
Ground state energy, E₁ =-13.6 eV.
So the energy of intermediate state n is,
Eₙ =-0.38 -1.13 eV
=-1.51 eV
(a) So the energy of the photon emitted in the second transition
=Eₙ -E₁
=-1.51 -(-13.6) eV
=12.1 eV.
(b) Since Eₙ =-13.6/n²
→n² =-13.6/(-1.51) =9
→n =3.
15. What is the energy of a hydrogen atom in the first excited state if the potential energy is taken to be zero in the ground state?
ANSWER: The kinetic energy of the electron in the nth orbit is,
K =mZ²e⁴/(8εₒ²h²n²)
In the ground state n =1, and for hydrogen Z =1. So the kinetic energy of the electron in the ground state,
K =me⁴/(8εₒ²h²)
=13.6 eV, when we put the respective values.
Given that the potential energy in the ground state, V =0.
So the total energy of the hydrogen atom in the ground state,
E₁ =K +V =13.6 eV. ------- (i)
In the first excited state, n =2. We know that the energy of state n =2 is higher by 10.2 eV than the state n =1. {When we take the P.E. zero when the nucleus and electron are widely apart, E₁ =-13.6 eV and E₂ =-3.4 eV. So E₂ -E₁ =-3.4 -(-13.6) eV =10.2 eV}
Hence the energy of the hydrogen atom in the first excited state, {from -(i)}
=E₁ +10.2 eV
=13.6 +10.2 eV
=23.8 eV.
16. A hot gas emits radiation of wavelength 46.0 nm, 82.8 nm, and 103.5 nm only. Assume that the atoms have only two excited states and the difference between consecutive energy levels decreases as energy is increased. Taking the energy of the highest energy state to be zero, find the energies of the ground state and the first excited state.
ANSWER: Enengy corresponds to the photon of wavelength 46.0 nm
=hc/𝜆
=6.63x10⁻³⁴*3x10⁸/46x10⁻⁹ J
=4.32x10⁻¹⁸ J
=27 eV.
The energy corresponding to the photon of wavelength 82.8 nm
=27*(46/82.8) eV
=15 eV
The energy corresponding to the photon of wavelength 103.5 nm
=27*(46/103.5) eV
=12 eV.
(i) Since the atom has only two excitation states, the photon corresponding to the radiation with energy 27 eV is for the transition of an electron from n =3 to n =1. Thus there is a difference of 27 eV of energy between n =1 and n=3. See the figure below,
The figure for Q-16
(ii) Given that the difference between energy levels decreases as energy is increased. So the difference in energy between n =1 and n =2 is 15 eV and the difference of energy between n =2 and n =3 is 12 eV.
If we take the energy of n =3 equal to zero then the energy of n =1 state (ground state) is
=0 -27 eV =-27 eV.
{Refer para (i) above}
Similarly, the energy of n =2 state (First excitation state) is
=0 -12 eV
=-12 eV.
{Refer para (ii) above}
17. A gas of hydrogen-like ions is prepared in a particular excited state A. It emits photons having a wavelength equal to the wavelength of the first line of the Lyman series together with photons of five other wavelengths. Identify the gas and find the principal quantum number of the state A.
ANSWER: Since there are six different wavelengths of emitted photons, there will be six different transitions. If an atom is excited to the principal quantum number n =m state then there will be m(m-1)/2 number of possible transitions. So here,
m(m-1)/2 =6
→m =4.
Hence the principal quantum number of state A is 4.
If Z is the atomic number of the hydrogen-like ion, the energy of the nth state,
E =-13.6Z²/n²
If there is a transition from state m to n where m > n, then the energy of the emitted photon
→E =-13.6Z²/m²-(-13.6Z²/n²) eV
=13.6Z²{1/n²-1/m²} eV ---- (i)
Given that the wavelength related to one group of photons is equal to the first line of the Lyman series. The first line of the Lyman series is obtained when there is a transition of an electron from state n =2 to n =1 in a hydrogen atom. It has an energy of
13.6{1-1/2²} eV =3*3.4 eV =10.2 eV.
Equating energy in (i) to it we get,
13.6Z²(1/n² -1/m²) =10.2
→Z² =0.75m²n²/(m²-n²)
Since atomic number Z will be a whole number, we try six different transition combinations of (m,n) to check what value is a whole number. The combinations will be (4,1), (4,2), (4,3), (3,1), (3,2), and (2,1).
We find that two combinations (4,2) and (2,1) give whole numbers for Z.
For (4,2)→Z² =0.75*64/12 =4
→Z =2.
Which is for a Helium ion.
For (2,1) →Z² =0.75*4/3 =1
→Z =1.
Which is for the hydrogen gas and the transition relates to the first line of the Lyme series itself. Since in the given problem, the gas in question is a hydrogen-like ion. Thus we take Z =2 for which the gas is identified as He⁺.
18. Find the maximum angular speed of the electron of a hydrogen atom in a stationary orbit.
ANSWER: The angular speed of the electron in a hydrogen atom is
⍵ =v/r
Since angular momentum is
mvr =nh/2π
→v =nh/(2πmr)
Putting this value of v in angular speed, we get
⍵ =nh/(2πmr²) --------------- (i)
We see that the angular speed is directly proportional to 'n' but inversely proportional to the square of the radius. Thus the maximum angular speed of the electron of a hydrogen atom will be for n =1 where the value of radius is minimum i.e. r =0.053 nm. Putting the values in (i) we get,
⍵ₘₐₓ=6.63x10⁻³⁴/{2π*9.1x10⁻³¹*(0.053x10⁻⁹)²} rads/s
=4.1x10¹⁶ rads/s
19. A spectroscope instrument can resolve two nearby wavelengths 𝜆 and 𝜆+Δ𝜆 if 𝜆/Δ𝜆 is smaller than 8000. This is used to study the spectral lines of the Balmer series of hydrogen. Approximately how many lines will be resolved by the instrument?
Note: This problem is also explained in a video, Click this link to view it.
ANSWER: Given 𝜆/Δ𝜆 < 8000.
For the Balmer series, the wavelength is given as
1/𝜆 =R{1/4 -1/m²}
Taking the differential on both sides, we get
-(1/𝜆²)*Δ𝜆 =(R*2/m³)Δm
→-(1/𝜆)*(Δ𝜆/𝜆)=2R*Δm/m³
→-R{1/4-1/m²}*(Δλ/λ) =2R*Δm/m³
→-(m²-4)(Δλ/λ)/4m² =2Δm/m³
The negative sign is for the fact that when m increases λ decreases. So we neglect it here. Also for the consecutive orbits, we take Δm =1. So now
(m²-4)m/8 =(λ/Δλ) <8000
By trial, for m =40, LHS =7980<8000
For m =41, LHS=8595 >8000
So this inequality is satisfied for m =40.
In this case of the Balmer series, there will be 38 transitions from m =40 to n=2 and only these 38 spectral lines will be resolved by the instrument.
20. Suppose, in certain conditions, only those transitions are allowed to hydrogen atoms in which the principal quantum number n changes by 2. (a) Find the smallest wavelength emitted by hydrogen. (b) List the wavelengths emitted by hydrogen in the visible range (380 nm to 780 nm).
ANSWER: (a) Smallest wavelength means the greatest energy. The energy of an orbit is inversely proportional to n². So higher value of n, the curve of energy E becomes very close and asymptotic to the n-axis. Therefore the energy of a transition is more in orbits near the nucleus for the same difference of n. Therefore here we shall consider the transition of an electron from n =3 to n =1. Hence this wavelength will be given as,
1/𝜆 =R{1/1² -1/3²} =8R/9
→𝜆 =9/8R
=9/{8*1.097x10⁷}
=1.03x10⁻⁷ m
=103x10⁻⁹ m.
=103 nm.
(b) The visible range lies in the Balmer series for which all the transitions are up to n =2. So in the given case, only the transition from n =4 to n =2 is considerable. The wavelength is
1/𝜆 =R{1/2² -1/4²}
=3R/16
→𝜆 =16/3R
=16/{3*1.097x10⁷} m
=4.86x10⁻⁷ m
=486x10⁻⁹ m
=486 nm.
21. According to Maxwell's theory of electrodynamics, an electron going in a circle should emit radiation of frequency equal to its frequency of revolution. What should be the wavelength of the radiation emitted by a hydrogen atom in the ground state if this rule is followed?
ANSWER: We know that the angular momentum of the electron is,
mvr =nh/2π.
Here n =1 and the radius of the orbit in this ground state r =0.053 nm.
So, mvr =h/2π.
Angular speed ⍵ =v/r
→⍵ =mvr/mr² =h/2πmr²
The frequency of revolution f is given as,
⍵ =2πf.
→f =⍵/2π
Since the frequency of radiation is equal to the frequency of revolution, f = c/λ, where λ is the wavelength and c is the speed of light.
Hence λ =c/f
=2πc/⍵
=2πc*2πmr²/h
=4π²mr²c/h
=4π²*9.1x10⁻³¹*(0.053x10⁻⁹)²*3x10⁸/6.63x10⁻³⁴ m
=4.57x10⁻⁸ m
=45.7x10⁻⁹ m
=45.7 nm.
22. The average kinetic energy of molecules in a gas at temperature T is 1.5kT. Find the temperature at which the average kinetic energy of the molecules of hydrogen equals the binding energy of its atom. Will hydrogen remain in its molecular form at this temperature? Take k = 8.62x10⁻⁵ eV/K.
ANSWER: The binding energy of a hydrogen atom =13.6 eV.
The average kinetic energy of molecules of a gas at temperature T is given equal to 1.5kT.
From the given condition,
1.5kT =13.6, putting the value of k from the given problem,
→1.5*8.62x10⁻⁵*T =13.6
→T =1.05x10⁵ K.
This is a very very high temperature and the hydrogen will not remain in its molecular form at this temperature.
23. Find the temperature at which the average thermal kinetic energy is equal to the energy needed to take a hydrogen atom from its ground state to n =3 state. Hydrogen can now emit red light of wavelength 653.1 nm. Because of the Maxwellian distribution of speed, a hydrogen sample emits red light at temperatures much lower than that obtained from this problem. Assume that hydrogen molecules dissociate into atoms.
ANSWER: Energy required to take a hydrogen atom from ground state to n =3 state is,
E =13.6{1/1² -1/3²} eV
=13.6*8/9 eV
=12.09 eV.
The average thermal kinetic energy is given as,
K.E. =1.5kT eV
Equating both the energies as per the given condition,
1.5kT =12.09
→T =12.09/1.5k
=12.09/(1.5*8.62x10⁻⁵) K
=9.4x10⁴ K.
24. The average lifetime of a hydrogen atom excited to n = 2 state is 10⁻⁸ s. Find the number of revolutions made by the electron on average before it jumps to the ground state.
ANSWER: Angular momentum of the electron of a hydrogen atom at n =2,
mvr =nh/2π =h/π
Angular speed ⍵ =v/r =mvr/mr²
→⍵ =h/πmr²
Frequency f is given as,
⍵ =2πf
→f =⍵/2π
=h/{2π²mr²}
=h/{2π²m(n²aₒ)²}, here n =2, so
f =h/{32π²maₒ²}
=6.63x10⁻³⁴/{32π²*9.1x10⁻³¹*(0.053x10⁻⁹)²} Hz
=8.21x10¹⁴ Hz
Hence the number of revolutions made by the electron in the average lifetime of t =10⁻⁸ s is
=f*t
=8.21x10¹⁴*10⁻⁸
=8.21x10⁶.
25. Calculate the magnetic dipole moment corresponding to the motion of the electron in the ground state of a hydrogen atom.
ANSWER: Current in the orbit (ground state) due to a revolving electron,
i =Q/t =e/{2πaₒ/v}
=ev/{2πaₒ}
But v =Ze²/(2εₒhn)
Putting Z =1 and n =1, we have
v =e²/(2εₒh)
Now, i =e³/{4πεₒhaₒ}
Hence the magnetic dipole moment,
=i*A
=(1/4πεₒ)*{e³/(haₒ)}*(πaₒ²)
=9x10⁹*(1.6x10⁻¹⁹)³*πaₒ/h
=1.16x10⁻⁴⁶*0.053x10⁻⁹/6.63x10⁻³⁴ A-m²
=9.2x10⁻²⁴ A-m².
26. Show that the ratio of the magnetic dipole moment to the angular momentum (l =mvr) is a universal constant for hydrogen-like atoms and ions. Find its value.
ANSWER: As we have seen in the previous problem, the magnetic dipole moment is
µ =i*A
=(ev/2πr)*πr²
=evr/2
The angular momentum
l =mvr
Hence the ratio,
µ/l =evr/(2mvr)
=e/2m =½(e/m)
Since e/m is the ratio of charge to mass of an electron which is a universal constant, hence µ/l =½(e/m) is also a universal constant. Its value is
=½*(1.6x10⁻¹⁹/9.1x10⁻³¹) C/kg
=8.8x10¹⁰ C/kg.
27. A beam of light having wavelengths distributed uniformly between 450 nm to 550 nm passes through a sample of hydrogen gas. Which wavelength will have the least intensity in the transmitted beam?
ANSWER: The given range of wavelengths lies in the visible range. Hence photons having energies equal to the transition of electrons from n >2 to n =2 may be absorbed by the hydrogen gas and these absorbed light wavelengths will have the least intensity in the transmitted beam.
The wavelength for transition n =3 to n =2,
1/λ =R{1/4 -1/9} =5R/36
→λ =656 nm
For the transition from n =4 to n=2,
1/λ =R{1/4 -1/16} =3R/16
→λ =486 nm
For the transition n =5 to n =2,
1/λ =R{1/4 -1/25} =21R/100
→λ =434 nm.
So only wavelength λ =486 nm of the given light beam (450 nm to 550 nm) may be absorbed by the hydrogen and this wavelength will have the least intensity in the transmitted beam.
28. Radiation coming from transitions n =2 to n =1 of hydrogen atoms fall on helium ions in n =1 and n =2 states. What are the possible transitions of helium ions as they absorb energy from the radiation?
ANSWER: The energy of radiation coming from transitions n =2 to n =1 of hydrogen atoms is
E =13.6*{1 -1/2²} =10.2 eV.
The energy required by an electron in helium ion for the transition from n =1 to n =2 states,
E' =Z²*13.6*{1 -1/4} eV
=40.8 eV.
It is more than 10.2 eV. Hence the photons coming from the transition in hydrogen atoms will not be absorbed by the helium ions for the transition from n =1 to n =2 states. All other transitions in helium ion from n = 1 state to higher states will require more than 40.8 eV of energy hence the coming photon will not be absorbed. Now consider the electron in the n = 2 state of the helium ion. The energy required by it for transition to n =3 states are,
E" =Z²*13.6{1/4 -1/9} =4*1.89 eV
=7.6 eV.
For transition from n =2 to n =4 states E* =4*13.6{1/4 -1/16} =10.2 eV.
For the next higher transitions energy required will be more than 10.2 eV.
The coming photons have energy =10.2 eV hence they may be absorbed by the helium ions for transitions from n = 2 to n =3 states or from n =2 to n =4 states that require energy less than or equal to 10.2 eV.
29. A hydrogen atom in the ground state absorbs a photon of ultraviolet radiation of wavelength 50 nm. Assuming that the entire photon energy is taken up by the electron, with what kinetic energy will the electron be ejected?
ANSWER: The energy of the photon of wavelength 50 nm is
E = hc/λ
=(4.14x10⁻¹⁵ eV-s)*3x10⁸ m/50x10⁻⁹ m
=24.8 eV
An electron in the ground state of a hydrogen atom requires 13.6 eV of energy to widely separate it from the nucleus. If this photon is fully absorbed by the electron then the rest of the energy will be conserved as the kinetic energy of the electron. Hence the electron will be ejected with a kinetic energy of 24.8 -13.6 eV =11.2 eV.
30. A parallel beam of light of wavelength 100 nm passes through a sample of atomic hydrogen gas in the ground state. (a) Assume that when a photon supplies some of its energy to a hydrogen atom, the rest of the energy appears as another photon moving in the same direction as the incident photon. Neglecting the light emitted by the excited hydrogen atoms in the direction of the incident beam, what wavelengths may be observed in the transmitted beam? (b) A radiation detector is placed near the gas to detect radiation coming from perpendicular to the incident beam. Find the wavelengths of radiation that may be detected by the detector.
ANSWER: (a) The energy of photons in the parallel beam,
E = hc/λ
(Since λ =100 nm)
→E =4.14x10⁻¹⁵*3x10⁸/100x10⁻⁹ eV
=12.42 eV
Since it is not necessary that this incident photon will essentially give its energy wholly or partly, a portion of this parallel beam will pass through the hydrogen gas intact. So in the transmitted beam original 100 nm wavelength will be observed.
Suppose an electron in the ground state absorbs energy to jump into the next state n =2, the absorbed energy
E =13.6{1 -1/2²} =10.2 eV
As per the given condition, the rest of the energy appears as another photon. Its energy
=12.42 -10.2 =2.22 eV
The wavelength of this observed photon
λ =hc/E
=4.14x10⁻¹⁵*3x10⁸/2.22 m
=5.60x10⁻⁷ m
=560x10⁻⁹ m
=560 nm.
If the electron jumps to n =3 state from the ground state, the absorbed energy
E =13.6{1 -1/9} =12.1 eV
The energy of the residual photon
=12.42 -12.1 =0.32 eV
Hence the observed wavelength of the transmitted beam
λ =4.14x10⁻¹⁵*3x10⁸/0.32 m
=38.80x10⁻⁷ m
=3880x10⁻⁹ m
=3880 nm.
If the electron tries to jump to the n =4 state from the ground state, the required energy for absorption
=13.6{1 -1/16} =12.75
but it is more than the incident photon. So it will not be absorbed.
(b) Since in the above-excited states the electrons have jumped to n =3 and n =2 states. They will ultimately revert back to the ground state either directly or in steps. So possible transitions are, from m =3 to n=1, n =2 to n =1 and n =3 to n =2.
The wavelengths observed in these transitions will be:-
For n =3 to n =1
1/λ =R{1 -1/9} =8R/9
→λ =9/8R =1.03x10⁻⁷ m
=103x10⁻⁹ m
=103 nm.
For n =2 to n =1 transition
1/𝜆 =R{1 -1/4} =3R/4
→𝜆 =4/3R =1.21x10⁻⁷ m
=121x10⁻⁹ m
=121 nm.
For n =3 to n =2 transition,
1/λ =R{1/4 -1/9} =5R/36
→λ =36/5R =6.56x10⁻⁷ m
=656x10⁻⁹ m
=656 nm.
31. A beam of monochromatic light of wavelength 𝜆 ejects photoelectrons from a cesium surface (Ჶ =1.9 eV). These photoelectrons are made to collide with hydrogen atoms in the ground state. Find the maximum value of 𝜆 for which (a) hydrogen atoms may be ionized, (b) hydrogen atoms may get excited from the ground state to the first excited state and (c) the excited hydrogen atoms may emit visible light.
ANSWER: (a) To ionize a hydrogen atom in the ground state, 13.6 eV of energy is needed. The emitted electrons from the cesium surface must possess 13.6 eV of kinetic energy. Due to the work function of the cesium surface, the minimum energy of the photons of the monochromatic light must be equal to 13.6 eV +1.9 eV =15.5 eV and in this case, the 𝜆 will have the maximum value.
Thus, for the maximum wavelength 𝜆, the energy of photons of monochromatic light should be
hc/𝜆 =15.5 eV
→𝜆 =hc/15.5
=4.14x10⁻¹⁵*3x10⁸/15.5 m
=8.0x10⁻⁸ m
=80x10⁻⁹ m
=80 nm.
(b) To get excited from the ground state to the first excitation state, a hydrogen atom needs energy equal to
E =13.6(1 -1/2²)
=10.2 eV
So the emitted electrons from the cesium surface need a minimum kinetic energy of 10.2 eV. In this case, the maximum value of 𝜆 i.e. minimum energy of the photons should be equal to 10.2 eV +1.9 eV =12.1 eV.
So, hc/𝜆 =12.1
→𝜆 =hc/12.1
=4.14x10⁻¹⁵*3x10⁸/12.1 m
=1.02x10⁻⁷ m
=102x10⁻⁹ m
=102 nm.
(c) The excited hydrogen atom emits visible light when electrons jump down to the n =2 state from higher states. So, the minimum kinetic energy of the electrons emitted from the cesium surface should be such that it takes the electron in the ground state of the hydrogen atom to n =3 state. This minimum energy =13.6{1 -1/3²) eV
=12.1 eV
So the energy of the photons of the monochromatic light should be at least
=12.1+1.9 eV
=14.0 eV
Hence
hc/𝜆 =14.0
→𝜆 =4.14x10⁻¹⁵*3x10⁸/14.0 m
=8.9x10⁻⁸ m
=89x10⁻⁹ m
=89 nm.
32. Electrons are emitted from an electron gun at almost zero velocity and are accelerated by an electric field E through a distance of 1.0 m. The electrons are now scattered by an atomic hydrogen sample in the ground state. What should be the minimum value of E so that red light of wavelength 656.3 nm may be emitted by the hydrogen?
ANSWER: The wavelength of the given red light
𝜆 =656.3 nm.
The energy of a photon of this wavelength,
E' =hc/𝜆
=4.14x10⁻¹⁵*3x10⁸/656.3x10⁻⁹ eV
=1.9 eV.
The red light is in the visible range and visible range lights are emitted in a hydrogen atom when the electrons from higher states come down to the n =2 states. Since an electron in the ground state requires energy =13.6*(1-1/2²) =10.2 eV to reach the n =2 state, the accelerated electrons must possess 10.2 +1.9 =12.1 eV energy to make the electrons in the ground state to jump to the required excited state.
Since the electrons are accelerated through 1.0 m in an electric field E, the work done on it (=eE joules =E eV) will be the kinetic energy of the electrons. Equating it with the energy needed we get,
E =12.1 V/m
33. A neutron having a kinetic energy of 12.5 eV collides with a hydrogen atom at rest. Neglect the difference in mass between the neutron and the hydrogen atom and assume that the neutron does not leave its line of motion. Find the possible kinetic energies of the neutron after the event.
ANSWER: We know that when two bodies of equal masses elastically collide in a linear direction, their velocities interchange. So the neutron in this case will have zero velocity after the collision and hence zero kinetic energy.
This can also be proved mathematically. The neutron does not leave its line of motion after the collision which means that the hydrogen atom at rest will also move along the same line after the collision (from the conservation of momentum in a direction principle).
Before collision let the momentum of the neutron =p and that of the hydrogen atom =0. After collision let the momenta of neutron = p' and that of hydrogen atom =p". So from the momentum conservation principle,
p+0 =p'+p"
→p' +p" =p ------------------ (i)
The kinetic energies of the neutron and hydrogen atom are K (=12.5 eV) and zero before the collision.
If the kinetic energies of neutron and hydrogen atoms are K' and K" respectively after the collision, then from the energy conservation rule
K'+K" =K +0.
Putting the expression of kinetic energy in terms of momentum
p'²/2M +p"²/2M =p²/2M
→p'² +p"² =p² ------ (ii)
But from (i)
(p' +p")² =p²
→p'² +p"² +2p'p" =p²
→p²+2p'p" =p² {from (ii)}
→2p'p" =0
→p'p" =0
After the collision, the hydrogen atom will have some velocity and momentum, p" ≠0 which requires p' to be zero.
So the neutron will have zero momentum after the collision.
34. A hydrogen atom moving at speed v collides with another hydrogen atom kept at rest. Find the minimum value of v for which one of the atoms may get ionized. The mass of a hydrogen atom =1.67x10⁻²⁷ kg.
ANSWER: Let after the collision, the velocities of hydrogen atoms are v' and v". From the momentum conservation principle,
Mv =Mv' +Mv"
→v =v' +v" -------- (i).
If the energy required for the ionization of one hydrogen atom is E, then from the conservation of energy principle,
½Mv² =½Mv'²+½Mv"²+E
→v² =v'²+v"²+2E/M
But by squaring both sides of (i), we get
v² =v'²+v"²+2v'v"
By comparison,
2E/M =2v'v"
→E/M =v'v"
Now,
(v'-v")² =(v'+v")²-4v'v"
→(v'-v")² = v² -4E/M
→v² =(v'-v")²+4E/M
For v to be minimum,
(v'-v")² =0
→v' =v"
That is both atoms move at the same speed after collision. Also, for the minimum v,
v² =4E/M
We know, E =13.6 eV. Hence,
v²=4*13.6x1.6x10⁻¹⁹/1.67x10⁻²⁷
=5.24x10⁹
→v =7.2x10⁴ m/s.
35. A neutron moving with a speed v strikes a hydrogen atom in ground state moving towards it with the same speed. Find the minimum speed of the neutron for which inelastic (completely or partially) collision may take place. The mass of neutron = mass of hydrogen =1.67x10⁻²⁷ kg.
ANSWER: From the conservation of momentum principle, both will have the same speed v after the collision but opposite in direction. It can be shown below. Let after the collision the speed of the neutron is v' and that of the hydrogen atom be v". Then from the momentum conservation principle,
Mv -Mv =Mv'+Mv"
→v' =-v"
So both have equal and opposite speeds.
Suppose an energy E is absorbed by the hydrogen atom in the collision, then from the conservation of energy principle,
½Mv² *2=½Mv'²*2 +E
→v² =v'² +E/M
For v to be minimum, v' =0 which means that after the collision both come to rest and,
v² =E/M
Now the minimum amount of energy E that a hydrogen atom in the ground state can absorb is the energy required for the transition from n =1 to n =2 state i.e. equal to 10.2 eV.
So, v² =10.2*1.6x10⁻¹⁹/1.67x10⁻²⁷
=9.77x10⁸ m²/s²
→v =3.13x10⁴ m/s.
36. When a photon is emitted by a hydrogen atom, the photon carries momentum with it. (a) Calculate the momentum carried by the photon when a hydrogen atom emits light of wavelength 656.3 nm. (b) With what speed does the atom recoil during this transition? Take the mass of the hydrogen atom =1.67x10⁻²⁷ kg. (c) Find the kinetic energy of the recoil of the atom.
ANSWER: (a) We have the wavelength 𝜆 =656.3 nm.
The momentum of this photon,
p =h/𝜆, where h is the plank's constant.
=6.63x10⁻³⁴/656.3x10⁻⁹ kg-m/s
=1.0x10⁻²⁷ kg-m/s.
(b) From the conservation of momentum principle, the same amount of momentum will be gained by the atom but in the opposite direction. Let the final speed of the atom =v. Then,
Mv =p
→v =p/M
=1.0x10⁻²⁷/1.67x10⁻²⁷ m/s
=0.60 m/s.
(c) The kinetic energy of the recoiled atom,
=½Mv²
=½*1.67x10⁻²⁷*(0.6)² J
=3.0x10⁻²⁸ J
=1.9x10⁻⁹ eV.
37. When a photon is emitted from an atom, the atom recoils. The kinetic energy of the recoil and the energy of the photon comes from the difference in energies between the states involved in the transition. Suppose, a hydrogen atom changes its state from n =3 to n =2. Calculate the fractional change in the wavelength of the light emitted, due to the recoil.
ANSWER: When we do not consider the recoil of the atom, the wavelength is given as,
1/𝜆 =1.097x10⁷{1/2² -1/3²}
→1/𝜆 =1.52x10⁶
→𝜆 =6.563x10⁻⁷ m
=656.3x10⁻⁹ m
=656.3 nm.
When we consider the recoil of the atom, the momentum gained in recoil by the atom is equal to the momentum of the photon in magnitude. This momentum,
p =h/𝜆
=6.63x10⁻³⁴/656.3x10⁻⁹ kg-m/s
=1.0x10⁻²⁷ kg-m/s
Hence the kinetic energy gained by the atom in recoil is
E =½Mv² =½(Mv)²/M
=p²/2M
=(1.0x10⁻²⁷)²/(2*1.67x10⁻²⁷) J
=3.0x10⁻²⁸ J
=1.9x10⁻⁹ eV.
In this case, the photon will have less energy by this amount of E. This revised energy of the photon will be
E' =13.6{1/2² -1/3²} -E
=1.9 eV -1.9x10⁻⁹ eV
The wavelength of this light,
𝜆' =hc/E'
Hence the change in the wavelength
𝜆' -𝜆 ={hc/(1.9-1.9x10⁻⁹)}-hc/1.9
=hc[(1.9-1.9+1.9x10⁻⁹)/{1.9*(1.9-1.9x10⁻⁹)}]
=hc[1.9x10⁻⁹/(1.9*(1.9-1.9x10⁻⁹)}
=hc*1.0x10⁻⁹/{1.9(1-10⁻⁹)}
=hc*1.0x10⁻⁹/1.9
Because 1-10⁻⁹ is almost equal to 1.
Hence the fractional change in the wavelength
𝜆'-𝜆/𝜆 ={hc*1.0x10⁻⁹/1.9}/(hc/1.9)
=1.9x10⁻⁹/1.9
=10⁻⁹.
38. The light emitted in the transition n =3 to n =2 in hydrogen is called Hₐ light. Find the maximum work function a metal can have so that Hₐ light can emit photoelectrons from it.
ANSWER: For Hₐ light to emit photoelectrons from a metal surface, its photons should have energy equal to or more than the work function. Hence the maximum work function = energy of the photon of Hₐ light,
=13.6*{1/2² -1/3²} eV
=1.9 eV.
39. Light from the Balmer series of hydrogen is able to eject photoelectrons from a metal. What can be the maximum work function of the metal?
ANSWER: Light from the Balmer series of hydrogen is emitted when electrons from higher states jump down to the n =2 state. The maximum work function of the metal from which this light is able to eject photoelectrons will be equal to the maximum energy of the photons. The maximum energy of a photon will be when the electron in the hydrogen atom jumps from n =∞ state down to n = 2 states. So the maximum work function,
φ =13.6{1/2² -1/∞²} eV
=13.6/4 eV
=3.4 eV.
40. Radiation from a hydrogen discharge tube falls on a cesium plate. Find the maximum possible kinetic energy of the photoelectrons. The work function of cesium is 1.9 eV.
ANSWER: The maximum energy of a photon of radiation from the hydrogen discharge tube will be when the emission is from the transition of the electron from n =∞ to n =1 state in the hydrogen atom. The energy of this photon will be,
E =13.6{1/1² -1/∞²} eV
=13.6 eV.
The work function of cesium φ =1.9 eV. Hence out of 13.6 eV of energy of the photon, 1.9 eV will be used to free the photoelectron from the metal surface and the rest will appear as the kinetic energy of the photoelectron. Hence the kinetic energy of the photoelectron. Hence the maximum possible kinetic energy of a photoelectron in this case is
=13.6 eV -1.9 eV
=11.7 eV.
41. A filter transmits only radiation of wavelength greater than 440 nm. Radiation from a hydrogen discharge tube goes through such a filter and is incident on a metal of work function 2.0 eV. Find the stopping potential that can stop the photoelectrons.
ANSWER: The wavelength around 440 nm falls in the Balmer series. The photons just more than 440 nm will have the maximum energy and when they eject photoelectrons their stopping potential will be maximum. Let 𝜆 be the wavelength of such radiation. In the hydrogen atom electrons from higher states jump down to the n =2 state. Let us put the wavelength =440 nm as a trial.
1/(440 nm) =1.097x10⁷{1/2² -1/n²}
→1/4 -1/n² =1/{(440x10⁻⁹*1.097x10⁷) →1/n² =1/4 -1/4.83 =0.043
→n² =23.3
→n =4.82
But n should be an integer and for the wavelength to be just more than 440 nm it should be less than 4.82. So the electron jumps down from n =4 state to n =2 states.
The energy of this photon
E =13.6{1/2² -1/4²}
=2.55 eV.
Given work function φ =2.0 eV
Let the stopping potential =Vₒ
Then from the Einstein equation,
E -φ =Vₒ
→Vₒ =2.55 eV -2.0 eV
=0.55 eV.
42. The earth revolves around the sun due to gravitational attraction. Suppose that the sun and the earth are point particles with their existing masses and that Bohr's quantization rule for angular momentum is valid in the case of gravitation. (a) Calculate the minimum radius the earth can have for its orbit. (b) What is the value of the principal quantum number n for the present radius? The mass of the earth = 6.0x10²⁴ kg, the mass of the sun =2.0x10³⁰ kg, earth-sun distance =1.5x10¹¹ m.
ANSWER: From Bohr's quantization rule, the possible orbits are those for which the angular momentum is an integral multiple of h/2π.
Let v be the speed of the earth in the nth orbit.
Then GMm/r² =mv²/r
→r =GM/v² ------------------- (i)
But from Bohr's quantization rule,
mvr =nh/2π
→v =nh/2πmr
Putting this expression in (i)
r =GM(2πmr)²/n²h²
→r =4π²GMm²r²/n²h²
→r =n²h²/(4π²GMm²)
(a) The minimum radius that can be from this rule will be obtained by putting n =1.
r =h²/(4π²GMm²)
=(6.63x10⁻³⁴)²/(4π²6.67x10⁻¹¹*2x10³⁰*(6x10²⁴)²} m
=2.3x10⁻¹³⁸ m.
(b) Present radius R =1.5x10¹¹ m
Clearly, R =rn²
→n² =R/r
=1.5x10¹¹/2.3x10⁻¹³⁸
=6.5x10¹⁴⁸
→n =2.5x10⁷⁴.
43. Consider a neutron and an electron bound to each other due to gravitational force. Assuming Bohr's quantization rule for angular momentum to be valid in this case, derive an expression for the energy of the neutron-electron system.
ANSWER: Let the mass of the neutron =M and the mass of the electron =m. If the radius of the orbit =r then,
GMm/r² =mv²/r, where v is the speed of the electron.
→v² =GM/r. ----------------- (i)
From Bohr's quantization principle,
mvr =nh/2π
→v =nh/2πmr
Putting this value of v in (i)
n²h²/4π²m²r² =GM/r
→r =n²h²/(4π²GMm²)
Now total energy of the system will be the sum of the kinetic energy and the potential energy. The kinetic energy,
K =½mv²
=½mGM/r
=½mGM*(4π²GMm²)/n²h²
=2π²G²M²m³/n²h²
The potential energy,
P =-GMm/r
=-GMm*(4π²GMm²/n²h²)
=-4π²G²M²m³/n²h²
Hence the total energy of the system,
=K+P
=2π²G²M²m³/n²h² -4π²G²M²m³/n²h²
=-2π²G²M²m³/n²h².
44. A uniform magnetic field B exists in a region. An electron projected perpendicular to the field goes in a circle. Assuming Bohr's quantization rule for angular momentum, calculate (a) the smallest possible radius of the electron (b) the radius of the nth orbit, and (c) the minimum possible speed of the electron.
ANSWER: (a) Force on the electron in the magnetic field =evB.
For the circular motion of the electron, the centripetal force =mv²/r
Equating the two,
mv²/r =evB
→r =mv/eB ------- (i)
From Bohr's quantiztion rule,
mvr =nh/2π
→mv =nh/2πr
evB=mv²/r; mv=eBr
From (i),
r =nh/(2πreB)
→r² =nh/(2πeB)
For the smallest possible radius, n =1. Putting n =1 in the above expression,
r² =h/(2πeB)
→r =√{h/(2πeB)}
(b) Since r² =nh/(2πeB), the radius of the nth orbit will be,
r =√{nh/(2πeB)}.
(c) From (i),
mv =erB
→v =erB/m
Putting the value of r,
v =(eB/m)*√{nh/(2πeB)}
=√[nhe²B²/(2πm²eB)]
=√[nheB/(2πm²)]
Except for n, all other entities are constant in this expression. Hence for the lowest possible speed of the electron n =1. It is equal to
v =√[heB/(2πm²)].
45. Suppose in an imaginary world the angular momentum is quantized to be an even integer multiple of h/2π. What is the longest possible wavelength emitted by hydrogen atoms in the visible range in such a world according to Bohr's model?
ANSWER: The visible range of wavelength is from 365.0 nm to 656.3 nm. Since the visible range is obtained from the transition of electrons from the higher state to the n =2 state. Let us check the state for the longest wavelength of the visible range i.e. 656.3 nm.
1/(656.3 nm) =1.097x10⁷{1/2² -1/n²}
→1/4 -1/n² =1/(1.097x10⁷*656.3x10⁻⁹)
→1/4 -1/n²=0.139
→1/n² =0.11
→n² =9
→n =3
But in the given imaginary world the angular momentum is quantized to be an even integer multiple of h/2π. So the state n =3 is not possible. So we will take the next higher even multiple, i.e. n =4. The longest wavelength 𝜆 is given as
1/𝜆 =1.097x10⁷{1/2² -1/4²}
=0.206x10⁷
→𝜆 =4.86x10⁻⁷ m
=486x10⁻⁹ m
=486 nm.
46. Consider an excited hydrogen atom in state n moving with a velocity v (v << c). It emits a photon in the direction of its motion and changes its state to a lower state m. Apply momentum and energy conservation principles to calculate the frequency 𝜈 of the emitted radiation. Compare this with frequency 𝜈ₒ emitted if the atom were at rest.
ANSWER: Though the velocity of hydrogen atom v << c, the emitted photon will have the velocity =c because it is electromagnetic radiation. The momentum of the photon p =E/c. Where E is the energy of the photon.
From the momentum conservation principle,
mv =mv' +p
→v' =v -p/m
From the energy conservation principle,
½mv² =½mv'² +E
→v² =v'² +2E/m
→v'² =v² -2E/m
→v² -p²/m² -2vp/m =v² -2E/m
→2E/m =p²/m² +2vp/m
→2E/m =E²/m²c² +2vE/mc
→E/mc² +2v/c =2
→E/mc² =2(1 -v/c)
→E =2mc²(1 -v/c) ------- (i)
When the hydrogen atom is at rest. Let the energy of the emitted photon =E'. Its momentum p' =E'/c.
From the momentum conservation principle,
mv' +p' =0
→mv' =-p'
→v' =-p'/m =-E'/mc
From the energy conservation principle,
½mv'² =E'
→v'² =2E'/m
→E'²/m²c² =2E'/m
→E' =2mc² --------- ----- (ii)
If 𝜈 is the frequency of the radiation when the hydrogen atom is moving and 𝜈ₒ the frequency of the radiation when the hydrogen atom is at rest then,
E =h𝜈 =2mc²(1 -v/c), from (i)
And E' =h𝜈ₒ =2mc², from (ii)
Hence,
E/E' =h𝜈/h𝜈ₒ =2mc²(1- v/c)/2mc²
→𝜈/𝜈ₒ =(1 -v/c)
→𝜈 =𝜈ₒ(1 -v/c).
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