Multiple Choice Questions on Bohr Atom Model

Questions on Bohr model of hydrogen-like atoms usually appear in medical and engineering and other degree entrance examinations. You can access all posts on Bohr model on this site by clicking on the label ‘Bohr model’ below this post. Here are a few more questions on Bohr model of hydrogen atom:

(1) The electron in a hydrogen atom in the ground state absorbs energy equal to 12.1 eV and gets elevated to the highest possible excited state. What will be change in the angular momentum of the electron? (h = Planck’s constant)

(a) h/π

(b) 2h/π

(c) 3h/π

(d) 4h/π

(e) h/2π

In the case of the hydrogen atom the energies of the electron are – 13.6 eV, – 3.4 eV, – 1.51 eV, – 0.85 eV etc. for values of n equal to 1, 2, 3, 4 etc. respectively. After absorbing 12.1 eV of energy, the electron in the innermost orbit having energy – 13.6 eV gets elevated to the 3^rd orbit where its energy is – 1.51 eV.

The angular momentum of the electron in the innermost orbit (of quantum no. n = 1) is h/2π. In the 3^rd orbit its angular momentum is 3h/2π. Therefore, the change in the angular momentum is (3h/2π) – (h/2π) = h/π

(2) If the ionisation potential of hydrogen atom is 13.6 volt, the energy required to remove an electron from the second orbit of hydrogen atom is

(a) 0.54 eV

(b) 0.85 eV

(c) 1.51 eV

(d) 3.4 eV

(e) 13.6 eV

By stating that the ionisation potential of hydrogen atom is 13.6 volt, you are informed that 13.6 eV of energy is required to remove an electron from the innermost orbit of the hydrogen atom. This is because of the energy –13.6 eV possessed by the electron in the innermost orbit. Since the electron in the 2^nd orbit possesses energy equal to –13.6/2² = –3.4 eV, the energy required to remove an electron from the second orbit of hydrogen atom is 3.4 eV.

(3) In the hydrogen atom the transition that gives radiation in the visible region is

(a) from n > 1 to n = 1

(b) from n > 1 to n = 1

(c) from n > 2 to n = 1

(d) from n > 3 to n = 1

(e) from n > 2 to n = 2

The hydrogen atom gives visible spectral lines in the balmer series because of the transitions from outer orbits to the 2^nd orbit. So the correct option is (e).

(4) In the hydrogen spectrum the frequency of a line resulting from the transition of the electron from the orbit of quantum number n_x to quantum number n₁ is f. In a hydrogen- like atom the same transition gives rise to a spectral line of frequency 9f. The hydrogen- like atom has atomic number

(a) 1

(b) 2

(c) 3

(d) 6

(e) 9

The energy (E_n) of the electron in the orbit of quantum number n in a hydrogen-like atom is given by

E_n = –13.6 z²/n² where z is the atomic number.

The energy difference between two states of the hydrogen-like atom is therefore z² times the energy difference in the case of the hydrogen atom. The frequency of the resulting spectral line also is z² times the frequency in the case of the hydrogen atom. Since the frequency is 9 times, the atomic number of the hydrogen like atom is 3.

[The hydrogen-like atom in this question is doubly ionised lithium (Li⁺⁺)].

You will find many useful questions in this section here as well as here.

Two Questions (MCQ) on X-rays

(1) The atomic numbers of the target materials in two X-ray tubes are in the ratio 1:3. The X-ray photon energies of the K_α lines of these materials are in the ratio

(a) 1:3 (b) 3:1 (c) 1:√3 (d) 1:9 (e) 1:1

According to Mosley’s law, the frequency ‘ν’ of a particular characteristic X-ray (such as K_α) is directly proportional to the square of the atomic number of the target. Since the photon energy is hν, it follows that the energy is directly proportional to the square of the atomic number. Therefore, the energies are in the ratio 1:9.

(2) The minimum wave length of X-rays produced by an X-ray tube operating at an anode voltage of 24.8 kV is very nearly

(a) 1.8 Ǻ (b) 1.5 Ǻ (c) 1 Ǻ (d) 0.8 Ǻ (e) 0.5 Ǻ

The minimum wave length X-ray photon will have the entire energy of the impinging electron, which is 24.8 keV.

Since the product of the photon energy in eV and the wave length in Angstrom is 12400, the minimum wave length is 12400 ÷ 24800 = 05 Ǻ.

[You may use the equation hc/λ = 24800 eV= 24800×1.6×10^–19 joule to calculate λ in metre after substituting for Planck’s constant ‘h’ and the speed of light ‘c’, but it will be time consuming].

Multiple Choice Questions on Matter Waves

The following MCQ on de Broglie waves appeared in Kerala Medical Entrance 2004 test paper:
An electron and a proton have the same de Broglie wave length. Then the kinetic energy of the electron is
(a) zero (b) infinity (c) equal to the kinetic energy of the proton (d) greater than the kinetic energy of the proton (e) none of these
Since the de Broglie wave length λ is given by λ = h/p where ‘h’ is Planck’s constant and ‘p’ is the linear momentum, we have p = h/λ. Therefore, the proton and the electron given in the question have the same linear momentum. The kinetic energy (E) is given by E = p²/2m. Since the mass (m) of the electron is less than that of the proton, it follows that the kinetic energy of the electron is greater [Option (d)].
Now consider the following MCQ:
A nucleus of mass ‘M’ at rest emits an α-particle of mass ‘m’. The de Broglie wave lengths of the α-particle and residual nucleus will be in the ratio
(a) m : M (b) (M+m) : m (c) M : m (d) √m:√M (e) 1 : 1
After the α-emission, the α-particle and the residual nucleus will fly off in opposite directions in accordance with the law of conservation of linear momentum. Since the momenta are equal in magnitude, the de Broglie wave lengths are the same and the ratio is 1:1 [Option (e)].
The following MCQ tests your basic knowledge regarding the electron orbits in the hydrogen atom:
The ratio of the de Broglie wave lengths of the electron in the first and the third orbits in the hydrogen atom is
(a) 1 : 1 (b) 1 : 3 (c) 1 : 9 (d) 1 : 6 (e) 1 : 27
You should remember that the orbit of quantum number ‘n’ is made of ‘n’ complete waves so that we have generally 2πr_n = nλ_n where r_nn is the radius of the nth orbit and λ_nn is the wave length of the electron in the n^th orbit. So, we have 2πr₁ = λ₁ for the first orbit and 2πr₃ = 3×λ₃ for the third orbit. Therefore λ₁/λ₃ = 3r₁/ r₃.
But the orbital radius ‘r’ is directly proportional to n². Therefore, λ₁/λ₃ = 3/9 = ⅓. The correct option is (b).
Now consider the following MCQ:
The kinetic energy of an electron is the same as that of a photon of wave length 3100 A.U. What is the wave length of this electron?
(a) 4 A.U. (b) 5.4 A.U. (c) 6.1 A.U. (d) 7.6 A.U. (e) 12.4 A.U.
In the case of a photon, the product λE = 12400 where the wave length λ is in Angstrom Unit (A.U.) and the energy E is in electron volt. Therefore, the energy of the photon of wave length 3100 A.U. is 12400/3100 = 4eV. Since the kinetic energy of the electron is the same as that of the photon, it follows that the kinetic energy of the electron is 4eV, which means this electron was accelerated by 4 volts. The wave length of an electron accelerated by ‘V’ volt is √(150/V) A.U. The answer to the problem is thus √(150/4) A.U. = √(37.5) = 6.1 A.U. approximately [Option (c)].
[You should remember that you can use the above simple relation for the wave length of an electron at small accelerating voltages only (in other words, at non-relativistic speeds only)].

Two Questions from Modern Physics

(1) The energy of a photon is 20eV. Its momentum in kg m/s is
(a) 2.56×10^-27 (b) 5.33×10^-27 (c) 1.066×10^-26 (d) 2.13×10^-26 (e) 3.18×10^-26
Since E=mc^2, momentum of the photon, p = E/c = (20×1.6×10^-19)/(3×10^8). Note that we have converted the energy in eV into joule. The answer is 1.066×10^-26 kg m/s given in option (c).
(2) The wave length associated with an electron having kinetic energy 6eV is
(a) 9A.U. (b) 5A.U. (c) 2.5A.U. (d) 1.5A.U. (e) 0.5A.U.
In the case of electrons, de Broglie wave length, λ = √ [ 150/V] A.U. where V is the accelerating voltage for the electron (= 6 volt since the energy is 6 eV).
Therefore, λ = √25 = 5 A.U.