Questions on Atomic Physics Including JEE (Advanced) 2013 Questions

“Happiness comes when your work and words are of benefit to others.”
– Buddha

Today we shall discuss a few multiple choice questions on atomic physics. Questions in this section are simple and interesting. You can work out most of the questions in this section without consuming much time and hence you will be justified in giving some preference to them. Here are the questions with their solution:

(1) An electron and a positron moving along a straight line in opposite directions with equal speeds suffer a head-on collision and get annihilated, producing two photons. Along which directions will the photons travel?

(a) Along straight lines at right angles

(b) Along straight lines inclined at 45 º

(c) Along straight lines inclined at 120 º

(d) Along a straight line in opposite directions

(e) Along straight lines arbitrarily oriented

The net momentum of the system consisting of an electron and a positron moving with the same speed in opposite directions is equal to zero since the electron and the positron have the same mass. Therefore, the net momentum of the products produced in the collision process must be zero in accordance with the law of conservation of momentum. Cancellation of the momenta of the photons (for achieving the condition of zero net momentum) is possible only if the photons travel along a straight line in opposite directions [Option (d)].

(2) Suppose the wave length of one of the photons produced in the pair annihilation process considered in the above question is λ. The wave length of the other photon is

(a) λ

(b) 2 λ

(c) 3 λ

(d) an integral multiple of λ

(e) any thing, which can not be theoretically predicted.

Since the net momentum of the system is zero, the photons must possess equal and opposite momenta. Momentum p of a photon is related to its energy E by

            p = E/c where c is the speed of light in free space.

The photons generated must therefore be of the same energy. In other words their wave lengths must be the same [Option (a)].

The following single correct answer type multiple choice question was included in the JEE (Advanced) 2013 question paper:

(3) A pulse of light of duration 100 ns is completely absorbed by a small object initially at rest. Power of the pulse is 30 mW and the speed of light is 3×10⁸ ms^–1. The final momentum of the object is

(a) 0.3×10^–17 kg ms^–1

(b) 1×10^–17 kg ms^–1

(c) 3×10^–17 kg ms^–1

(d) 9×10^–17 kg ms^–1

Since the light pulse is completely absorbed by the object, the entire momentum of the pulse is transferred to the object. If E is the energy of the pulse, the momentum p is given by

            p =  E/c where c is the speed of light in free space.

But E = Pt where P is the power and t is the duration of the pulse.

Therefore, we have

             p = Pt/c = (30×10^–3)×(100×10^–9) /( 3×10⁸) = 1×10^–17 kg ms^–1

The following single digit integer answer type question (in which the answer is an integer ranging from 0 to 9) also was included in the JEE (Advanced) 2013 question paper:

(4) The work functions of silver and sodium are 4.6 and 3.2 eV respectively. The ratio of the slope of the stopping potential versus frequency plot for silver to that of sodium is:

Ans : ?

The maximum kinetic energy KE_max of the photo electron is given by

            KE_max = hν – φ where h is Planck’s constant, ν is the frequency of the incident radiation and φ is the work function of the photo emitting surface.

If the stopping potential is V, we have

            KE_max = eV where e is the electronic charge.

Therefore, eV = hν – φ from which

            V = (h/e)ν – φ/e

The above equation shows that if the stopping potebtial V is plotted against the frequency ν, a straight line graph of slope h/e is obtained.

Since h/e is a constant, the slope is the same for all photo emitters. Therefore, the  ratio of  slopes = 1.

Thus Ans. = 1 

You will find some useful multiple choice questions from atomic physics and nuclear physics here as well as here.

AIEEE 2008- An Imaginary Question on Bohr Model

The following question was included in the AIEEE 2008 question paper:

Suppose an electron is attracted towards the origin by a force k/r where ‘k’ is a constant and ‘r’ is the distance of the electron from the origin. By applying Bohr’s model to this system, the radius of the n^th orbital of the system is found to be ‘r_n’ and the kinetic energy of the electron to be ‘T_n’. Then which of the following is true?

(1) T_n α 1/n², r_n α n²

(2) T_n independent of n, r_n α n

(3) T_n α 1/n, r_n α n

(4) T_n α 1/n, r_n α n²

The force k/r supplies the centripetal force for the circular motion of the electron so that we have

k/r = mv²/r where ‘m’ is the mass and ‘v’ is the speed of the electron.

Therefore, mv² = k which is independent of the quantum number ‘n’. The kinetic energy T_n of the electron is ½ mv² which is therefore independent of the quantum number ‘n’.

Also, v = √(k/m).

The angular momentum of the electron in the n^th orbit of radius r_n is mvr_n and in the Bohr model mvr_n = nh/2π where ‘h’ is Planck’s constant. Substituting for ‘v’ we have

m×√(k/m) ×r_n = nh/2π

This gives r_n α n. So the correct option is (2).

In the above question the attractive force on the electron was imagined to be inversely proportional to the distance just for the sake of testing your problem solving skill. In a real hydrogen atom the force is certainly inversely proportional to the square of the distance. You will find questions on real Bohr model on this site by clicking on the label ‘Bohr model ‘ below this post.

IIT-JEE 2008: Linked Comprehension Type Multiple Choice Questions on Bohr Model of Hydrogen-like Atoms

The IIT-JEE Physics question paper is no more a nightmare to students aspiring for a seat in one of the IIT’s, consequent on the changes effected from 2007. If you have good grasp of fundamentals, you should definitely take up the challenge for brightening your future. Students appearing for other entrance examinations such as Physics GRE, AP Physics, AIEEE, KEAM and the like also will find the IIT-JEE question paper very useful for their preparation.

Here are three multiple choice questions on Bohr atom model which appeared under Linked Comprehension Type in IIT-JEE 2008 question paper:

Paragraph for Question Nos. 1 to 3

In a mixture of H – He⁺ gas (He⁺ is singly ionized He atom) H atoms and He⁺ ions are excited to their respective first excited states. Subsequently, H atoms transfer their total excitation energy to He⁺ ions (by collisions). Assume that the Bohr model of atom is exactly valid.

(1) The quantum number n of the state finally populated in He⁺ ions is

(A) 2

(B) 3

(C) 4

(D) 5

The singly ionised helium atom is hydrogen like since it contains a single electron. In a hydrogen like atom the energy (E) of the electron in the n^th orbit is given by

E = – 13.6 Z²/n² electron volt where Z is the atomic number.

For hydrogen atom the energies are – 13.6 eV, – 3.4 eV, – 1.51 eV and – 0.85 eV for values of n equal to 1, 2, 3 and 4 respectively

For the helium ion the energies are four fold, equal to – 54.4 eV, – 13.6 eV, – 6.04 eV and – 3.4 eV for values of n equal to 1, 2, 3 and 4 respectively

The energy difference between the electrons in first excited states of hydrogen atom and helium ion is (– 3.4 eV) – (– 13.6 eV) = 10.2 eV.

On collision, the helium ion absorbs this energy and occupies the 3^rd excited state (n = 4)

of energy – 13.6 eV + 10.2 eV = –3.4 eV.

Therefore the correct option is (C).

(2) The wave length of light emitted in the visible region by He⁺ ions after collision with H atoms is

(A) 6.5×10^–7 m

(B) 5.6×10^–7 m

(C) 4.8×10^–7 m

(D) 4.0×10^–7 m

The visible photons have wave lengths in the range 4000 Ǻ to 7000 Ǻ and their energies lie in the range 3.1 eV to 1.771 eV.

[It will be useful to remember the range of energies in eV. You will definitely remember the wave length range of visible photons which you can convert into energy range using the equation, λ E = 12400  where λ is the wave length in Angstrom and E is the energy in electron volt].

The energy difference between the He⁺ ion levels for n = 3 and n = 4 correspond to visible photon energy: – 3.4 eV– (– 6.04 eV) =  2.64 eV

The wave length of the light emitted in the visible region is therefore 124000/2.64 = 4800 Ǻ, nearly. The answer is thus 4.8×10^–7 m [Option (C)].

(3) The ratio of the kinetic energy of the n = 2 electron for the H atom to that of the He⁺ ion is

(A) ¼

(B) ½

(C) 1

(D) 2

The kinetic energy of the electron in the n^th orbit is 13.6 Z²/n².

[Note that the kinetic energy and the total energy of the electron are equal in amount. But the kinetic energy is positive where as the total energy is negative].

The required ratio is Z_H²/Z_He² = ¼.

All posts on Bohr model of hydrogen atom on this site can be accessed by clicking on the label ‘Bohr model’ below this post.

You can find a few posts on hydrogen atom at apphysicsresources.blogspot.com also. One such post is here.

Two Questions (MCQ) on Bohr Atom Model

You can find the earlier questions (with solution) on Bohr atom model posted on this site by clicking on the label ‘Bohr model’ or ‘hydrogen atom’ below this post. You can get them also by using the search option at the top of this page. Today I give you two questions which are meant for gauging the depth of your understanding of Bohr’s theory.

(1) When the electron in a hydrogen atom of mass M undergoes transition from an orbit of higher quantum number n₂ to an orbit of lower quantum number n₁, the recoil velocity acquired by the atom is (Rydberg’s constant = R, Planck’s constant = h)

(a) (R/hM) (1/n₁² – 1/n₂²)

(b) (hR/M) (n₂ – n₁)

(c) 1/hRM (1/n₁² – 1/n₂²)

(d) h/RM) (1/n₁² – 1/n₂²)

(e) (hR/M) (1/n₁² – 1/n₂²)

The wave number of the photon emitted because of the electron transition is

ν' = 1/λ = R(1/n₁² – 1/n₂²) where λ is the wave length of the photon and R is Rydberg’s constant.

The momentum of the photon is p = h/λ = hR(1/n₁² – 1/n₂²) where h is Planck’s constant.

When the photon is emitted with this momentum, the atom recoils (like a gun firing a bullet) with an equal and opposite momentum. Therefore, the recoil velocity of the atom is given by

v = p/M = (hR/M)(1/n₁² – 1/n₂²).

(2) If the radius of the innermost electron orbit in a hydrogen atom is R₁, the de Broglie wave length of the electron in the second excited state is

(a) πR₁

(b) 3πR₁

(c) 4πR₁

(d) 6πR₁

(e) 9πR₁

The wave length of the electron in the n^th orbit is given by

λ = 2πR_n/n where R_n is the radius of the n^th orbit.4

[This follows because the angular momentum of the electron in the n^th orbit is

mvR_n = nh/2π.

Therefore, de Broglie wave length, λ = h/mv = 2πR_n/n ]

The second excited state has quantum number n = 3 (Third orbit). The radius of the 3^rd orbit in terms of the radius R₁ of the first orbit is given by (remembering R_n = n² R₁)

R₃ = 9R₁

Therefore, λ = 2πR_n/n = 2π×9R₁/3 = 6πR₁

[It will be convenient to remember that the de Broglie wave length of the electron in the n^th orbit is n times the the wave length in the innermost orbit].

You will find some useful posts on Atomic Physics and Quantum effects at apphysicsresources