“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*4**E*, its de Broglie wave length will be
(a)

*λ/*4
(b)

*λ/*2
(c)

*λ*
(d)

*2**λ*
(e)

*4**λ*
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 = p*^{2}/2*m*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 4*m*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*p*_{1}and*p*_{2}are the momenta of the particles A and B, on equating their kinetic energies, we have*p*

_{1}

^{2}/2

*m*=

*p*

_{2}

^{2}/(2×4

*m*)

[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

*p*

_{1}/

*p*

_{2}= 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

*λ*

_{1}/

*λ*

_{2}=

*p*

_{2}/

*p*

_{1}= 2, as given in option (b).

(3) A metallic
surface is found to emit photo-electrons when monochromatic light rays of
frequencies n

_{1}and n_{2}(n_{2}> n_{1}) 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) (3n

_{1 }– n_{2})/ 2
(b) (2n

_{1 }– n_{2})/ 2
(c) (3n

_{1 }– n_{2})/ 3
(d) (n

_{2 }– n_{1})/ 3
(e) (n

_{2 }– n_{1})/ 2
If the threshold
frequency is n

_{0}and the maximum values of kinetic energy in the two cases are*E*_{1}and*E*_{2}respectively, we have*h*n

_{1}=

*h*n

_{0}+

*E*

_{1}and

*h*n

_{2}=

*h*n

_{0}+

*E*

_{2}

Therefore,

*E*_{1}/*E*_{2}= (n_{1}– n_{0})/(n_{2}– n_{0})
Since the ratio is
1 : 3 we have

(n

_{1}– n_{0})/(n_{2}– n_{0}) = 1/3
Or, 3n

_{1 }– 3n_{0}= n_{2}– n_{0}
This gives n

_{0}= (3n_{1 }– n_{2})/ 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/

*Ze*^{2}
(b) 1/

*Ze*
(c)

*v*
(d)

*m*
(e) 1/

*v*^{2}
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

*mv*^{2}= (1/4πε_{0})(2*Ze*^{2}/*r*) where*r*is the distance of closest approach.
[Remember that the
charge on the alpha particle is 2

*e*].
This gives

*r*=*Ze*^{2}/πε_{0}*mv*^{2}
This shows that

*r*is directly proportional to 1/*v*^{2}.
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