Question

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

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

Answer 1

Hint: Energy of photon is given by:
$E=h\upsilon$ (h is planck's constant ,$\upsilon$ is the frequency of photon)
$\upsilon ={\displaystyle \frac{C}{\lambda }}$ (C is the velocity of light and $\lambda$ is the wavelength)
Using Einstein energy equation:
Energy of photon = kinetic energy-work function.
$K_e=E-{\varphi }_0$ (E, is the energy of a photon, Ke is the kinetic energy and ${\varphi }_0$is the work function wave.)
$K_e={\displaystyle \frac{1}{2}}m{v}^2$
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: $E={\displaystyle \frac{hC}{\lambda }}$ ($\upsilon ={\displaystyle \frac{C}{\lambda }}$)-----(b)
Kinetic energy is given by:
$K_e={\displaystyle \frac{1}{2}}m{v}^2$
Or
$K_e=E-{\varphi }_0$ -----(g)
We can write kinetic energy in terms of momentum by multiplying both numerator and denominator by m.
$\Rightarrow K_e={\displaystyle \frac{1}{2}}{\displaystyle \frac{m{v}^2\times m}{m\times m}}$
Momentum is denoted by p, where p is:
$p=mv$
Substituting the value of p in the expression of Kinetic energy:
  $\Rightarrow K_e={\displaystyle \frac{1}{2}}{\displaystyle \frac{p^2}{m}}$ -----(1)
Momentum multiplied by velocity gives us energy:
$E=pC$
 $\therefore pC={\displaystyle \frac{hC}{\lambda }}$
Substituting the value of p in equation 1
  $\Rightarrow K_e={\displaystyle \frac{1}{2}}{\displaystyle \frac{h^2}{m{{\lambda }_d}^2}}$-----(2)

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

$$\begin{gathered} \Rightarrow \left({\displaystyle \frac{1}{2}}{\displaystyle \frac{h^2}{m{{\lambda }_d}^3}}\right)d{\lambda }_d=\left({\displaystyle \frac{hC}{{\lambda }^2}}\right)d\lambda \\ \Rightarrow {\displaystyle \frac{d{\lambda }_d}{d\lambda }}={\displaystyle \frac{{{\lambda }_d}^3}{{\lambda }^2}} \end{gathered}$$

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 $f_0$.