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.

Two EAMCET (Medical) 2009 Questions on Photons and Electrons

Today we will discuss two multiple choice questions from modern physics. These questions are from the section ‘electrons and photons’ and they were included in the EAMCET (Medical) 2009 question paper:

(1) In Millikan’s oil drop experiment, a charged oil drop of mass 3.2×10^–14 kg is held stationary between two parallel plates 6 mm apart, by applying a potential difference of 1200 V between them. How many electrons does the oil drop carry?

(1) 7

(2) 8

(3) 9

(4) 10

The electric field ‘E’ between the plates is V/d where V is the potential difference and d is the separation between the plates.

Therefore E = 1200/(6×10^–3) = 2×10⁵ volt per metre.

The electric force on the drop is Eq where ‘q’ is the charge on the drop. Since the drop is stationary, the electric force balances the weight mg of the drop.

Therefore Eq = mg so that

q = mg/E = (3.2×10^–14×10)/( 2×10⁵) = 1.6×10^–18 coulomb.

Since the electronic charge is 1.6×10^–19 coulomb, the number of excess electrons carried by the drop is (1.6×10^–18) /(1.6×10^–19) = 10.

(2) Electrons accelerated by a potential of ‘V’ volts strike a target material to produce ‘continuous X-rays’. Ratio between the de Broglie wave length of the electrons striking the target and the shortest wave length of the ‘continuous X-rays’ emitted is

(1) h/√(2Vem)

(2) (1/c) √(2m/Ve)

(3) (1/c) √(Ve/2m)

(4) he/√(Ve/2m)

The de Broglie wave length λ of the electron is given by

λ = h/p where h is Planck’s constant and p is the momentum of the electron.

The shortest wave length λ_min of the ‘continuous X-rays’ produced by the striking electrons is given by

hc/λ_min = Ve where c is the speed of light in free space and e is the electronic charge.

[Note that we have equated the entire energy of the electron to the energy of the X-ray photon for obtaining minimum wave length (or, the maximum frequency ν_max):

eV = hν_max = hc/λ_min].

Therefore, λ_min = hc/Ve.

The ratio of wave lengths required in the problem is

λ /λ_min = (h/p)/(hc/Ve ) = Ve/pc………..(i)

The kinetic energy of the electron is eV and the momentum p of the electron is related to its kinetic energy as

eV = p²/2m where m is the mass of the electron.

Therefore, p = √(2mVe)

Substituting this in Eq (i), we obtain

λ /λ_min = (1/c) √(Ve/2m)