Question

A photoelectric material having work function ${\phi _0}$ is illuminated with light of wavelength $\lambda \left(\lambda < \dfrac{{{h_c}}}{{{\phi _0}}} \right)$.
The fastest photoelectron has a de Broglie wavelength${\lambda _d}$. A change in the wavelength of the incident light by $\Delta \lambda $ results in change $\Delta {\lambda _d}$in ${\lambda _d}$. Then the ration $\dfrac{{\Delta {\lambda _d}}}{{\Delta \lambda }}$ is proportional to:

a. $\dfrac{{{\lambda ^2}_d}}{{{\lambda ^2}}}$
b. $\dfrac{{{\lambda _d}}}{\lambda }$
c. $\dfrac{{{\lambda ^3}_d}}{\lambda }$
d. $\dfrac{{{\lambda ^3}_d}}{{{\lambda ^2}}}$

Answer 1

Hint: Energy of photon is given by:
$E = h\upsilon $ (h is planck's constant ,$\upsilon $ is the frequency of photon)
$\upsilon = \dfrac{C}{\lambda }$ (C is the velocity of light and $\lambda $ is the wavelength)
Using Einstein energy equation:
Energy of photon = kinetic energy-work function.
${K_e} = E - {\phi _0}$ (E, is the energy of a photon, Ke is the kinetic energy and ${\phi _0}$is the work function wave.)
${K_e} = \dfrac{1}{2}m{v^2}$
With the help of above relations we will solve the problem.

Complete step by step answer:
Work function: the minimum amount of energy required to eject an electron out of the metal surface is called the work function of the metal.
Photon: light consists of particles associated with a definite amount of energy and momentum .These particles were named as Photon.
Now, come to the mathematical part of the problem:
From the hint we have:
$E = h\upsilon $
Or we can write as: $E = \dfrac{{hC}}{\lambda }$ ($\upsilon = \dfrac{C}{\lambda }$)-----(b)
Kinetic energy is given by:
${K_e} = \dfrac{1}{2}m{v^2}$
Or
${K_e} = E - {\phi _0}$ -----(g)
We can write kinetic energy in terms of momentum by multiplying both numerator and denominator by m.
$ \Rightarrow {K_e} = \dfrac{1}{2}\dfrac{{m{v^2} \times m}}{{m \times m}}$
Momentum is denoted by p, where p is:
$p = mv$
Substituting the value of p in the expression of Kinetic energy:
  $ \Rightarrow {K_e} = \dfrac{1}{2}\dfrac{{{p^2}}}{m}$ -----(1)
Momentum multiplied by velocity gives us energy:
$E = pC$
 $\therefore pC = \dfrac{{hC}}{\lambda }$
Substituting the value of p in equation 1
  $ \Rightarrow {K_e} = \dfrac{1}{2}\dfrac{{{h^2}}}{{m{\lambda _d}^2}}$-----(2)

Equating equation (g) and (2) we have:
  $
   \Rightarrow \dfrac{1}{2}\dfrac{{{h^2}}}{{m{\lambda _d}^2}} = E - {\phi _0} \\
   \Rightarrow \dfrac{1}{2}\dfrac{{{h^2}}}{{m{\lambda _d}^2}} = \dfrac{{hC}}{\lambda } \\
$ -----(3)
Multiplying and dividing LHS and RHS by wavelengths and differentiating both LHS and RHS by wavelengths.

$
   \Rightarrow \left(\dfrac{1}{2}\dfrac{{{h^2}}}{{m{\lambda _d}^3}}\right)d{\lambda _d} = \left(\dfrac{{hC}}{{{\lambda ^2}}}\right)d\lambda \\
   \Rightarrow \dfrac{{d{\lambda _d}}}{{d\lambda }} = \dfrac{{{\lambda _d}^3}}{{{\lambda ^2}}} \\
 $

Hence, the correct answer is option (D).

Note: Kinetic Energy of the photon does not depend on the intensity of light wave rather it depends on the frequency of the light wave on the other hand Photoelectrons emitted per second does depend on the intensity of light wave and not on the frequency of the light. The frequency of light must be greater than the threshold frequency $f_0$.