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.

Questions on Bohr Model of Hydrogen Atom

Questions based on the Bohr model of hydrogen atom are inevitable in any Medical and Engineering test paper. Consider the following two questions which appeared in A.I.I.M.S.- 2005 test paper:
(1) Solid targets of different elements are bombarded by highly energetic electron beams. The frequency (f) of the characteristic X-rays emitted from different targets varies with atomic number Z as
(a) f α √Z (b) f α Z² (c) f α Z (d) f α Z^3/2
As you might have noted, X-rays are produced by electron transitions from outer orbits to inner orbits. But atoms of high atomic number are required for the production of X-rays since the X-ray photon has much greater energy compared to light photon. The energy of the electron in an orbit is directly proportional to Z². [Note that in a hydrogen like atom, the energy is -13.6 Z²/n² electron volt]. The energy difference between levels also is directly proportional to Z². Since the energy difference is equal to hν where ‘ν’ is the frequency of the radiation emitted, the correct option is (b).
[More rigorous treatment shows that ν α (Z-b)² where b is a constant for a given spectral series].
(2) The ground state energy of hydrogen atom is -13.6 eV. What is the potential energy of the atom in this state?
(a) 0 eV (b) -27.2 eV (c) 1eV (d) 2 eV
The correct option is (b) since the potential energy is twice the kinetic energy. You should note that in all cases of central field motion under inverse square law attractive force, the total energy and potential energy are negative and the potential energy is twice the total energy.
Suppose we modify this question as follows:
The ground state energy of hydrogen atom is -13.6 eV. What is the kinetic energy of the electron in this state?
(a) -13.6 eV (b) -27.2 eV (c) 0 eV (d) 13.6 eV

The correct option is (d) since the kinetic energy is always positive and its magnitude is the same as that of the total energy. This is true in all cases of central field motion under an attractive inverse square law force, as in the case of the motion of a satellite around a planet.

You will find more multiple choice questions (with solution) at physicsplus: Questions on Bohr Atom Model