“Happiness comes when your work and words
are of benefit to others.”
– Buddha
– 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 KEmax of the photo electron is given by
KEmax = 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
KEmax = 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.
