“Non-violence leads to the highest ethics,
which is the goal of all evolution. Until we stop harming all other living
beings, we are still savages.”
– Thomas A. Edison
The sections
‘atomic physics and quantum effects’ and ‘nuclear physics’ in the AP Physics B
syllabus will appear to be interesting to most of the AP Physics B aspirants.
Today we will discuss a few practice questions (multiple choice) in these
sections:
(1) The de Broglie
wave length of a particle with kinetic energy E is λ. If its kinetic energy is increased to 4E, its de Broglie wave length will be
(a) λ/4
(b) λ/2
(c) λ
(d) 2λ
(e) 4λ
Since the kinetic
energy is directly proportional to the square of momentum, the momentum of the
particle is doubled when its kinetic energy is quadrupled.
[Note that kinetic
energy E = p2/2m where p is the momentum and m is the mass].
The de Broglie wave
length λ is given by
λ = h/p where h
is Planck’s constant and p is the
momentum.
Therefore, when p is doubled, λ is halved [Option (b)].
(2) Particles A and B
have masses m and 4m respectively but they carry the same
charge. When they are accelerated by the same voltage, their de Broglie wave
lengths are in the ratio
(a) 1 : 1
(b) 2 : 1
(c) 4 : 1
(d) 1 : 4
(e) 1 : 8
Let V represent the common accelerating
voltage and q represent the common
charge of the particles. If p1
and p2 are the momenta of
the particles A and B, on equating their kinetic energies, we have
p12/2m = p22/(2×4m)
[Remember that the
kinetic energy of a particle of charge q accelerated
by a voltage V is qV. The kinetic energies of A and B are
equal since they have the same charge and they are accelerated by the same
voltage]
The above equation
gives
p1/ p2 = ½
Since the de
Broglie wave length λ is given by
λ = h/p where h
is Planck’s constant and p is the
momentum, the ratio of the de Broglie wave lengths of A and B is given by
λ1/ λ2
= p2/ p1
= 2, as given in option (b).
(3) A metallic
surface is found to emit photo-electrons when monochromatic light rays of
frequencies n1 and n2 (n2 > n1) are incident on it.
If the maximum values of kinetic energy of the photo-electrons emitted in the
two cases are in the ratio 1 : 3, the threshold frequency of the metallic
surface is
(a) (3n1 – n2)/ 2
(b) (2n1 – n2)/ 2
(c) (3n1 – n2)/ 3
(d) (n2 – n1)/ 3
(e) (n2 – n1)/ 2
If the threshold
frequency is n0 and the maximum
values of kinetic energy in the two cases are E1 and E2
respectively, we have
hn1 = hn0 + E1 and
hn2 = hn0 + E2
Therefore, E1/E2 = (n1 – n0)/(n2 – n0)
Since the ratio is
1 : 3 we have
(n1 – n0)/(n2 – n0) = 1/3
Or, 3n1 – 3n0 = n2 – n0
This gives n0 = (3n1 – n2)/ 2, as given in
option (a).
(4) The energy that
must be added to an electron to reduce its de Broglie wave length from 2 nm to
1 nm is
(a) half the initial energy
(b) equal to the initial energy
(c) twice the initial energy
(d) thrice the initial energy
(e) four times the initial energy
The de Broglie wave
length λ is given by
λ = h/p where h
is Planck’s constant and p is the
momentum.
Since the de Broglie wave length of the electron is to be
reduced to half the initial value
(from 2 nm to 1 nm), the momentum of the electron is to be doubled. But when the momentum is doubled, its kinetic energy
becomes four times the initial value.
Therefore the energy that must be added is three
times the initial energy
[Option (d)].
(5) An alpha particle
of mass m and speed v
proceeds directly towards a heavy nucleus of charge Ze. The distance of closest approach of the alpha particle is
directly proportional to
(a) 1/Ze2
(b) 1/Ze
(c) v
(d) m
(e) 1/v2
The alpha particle
has to move towards the nucleus with difficulty, doing work against the
electrostatic repulsive force. When the alpha particle reaches the distance of
closest approach, the entire kinetic energy gets converted into electrostatic
potential energy. Therefore we have
½ mv2 = (1/4πε0)(2Ze2/r) where r is the distance of closest approach.
[Remember that the
charge on the alpha particle is 2e].
This gives r = Ze2/πε0mv2
This shows that r is directly proportional to 1/v2.
