In a photoelectric experiment three different lights are incident on a metal of work function $1.5 \mathrm{eV}$. Light $A$ is $200 \mathrm{~nm}$ wavelength with intensity $1.8 \mathrm{~W} / \mathrm{m}^2$, light $B$ is $400 \mathrm{~nm}$ with $1 \mathrm{~W} / \mathrm{m}^2$ and light $C$ is $600 \mathrm{~nm}$ with $0.5 \mathrm{~W} / \mathrm{m}^2$. The current versus voltage is measured. Which graphs correspond to which light?

• A

  I III II
• B

  I II IV
• C

  I III IV
• Correct answer
  D

  III II IV

Solution

Given that, work-function, $\phi_0=1.5 \mathrm{eV}$ Let $\lambda_0=$ threshold wavelength. Then, $\phi_0=\frac{h c}{\lambda_0}$ $$ \begin{aligned} \Rightarrow \quad 1.5 \mathrm{eV} & =\frac{1242}{\lambda_0}(\mathrm{eV}-\mathrm{nm}) \\ \lambda_0 & =828 \mathrm{~nm} \end{aligned} $$ Now, photocurrent is directly proportional to intensity of incident light.
$V_1, V_2, V_3$ and $V_4$ are cut off voltage, which is required to stop the electron having maximum kinetic energy. Given that, $\lambda_A=200 \mathrm{~nm}$, its corresponding cut off voltage is $V_2$. $\lambda_B=400 \mathrm{~nm}$, its corresponding cut off voltage is $V_3$. $\lambda_C=600 \mathrm{~nm}$, its corresponding cut off voltage is $V_4$. and corresponding photocurrent, $$ \begin{aligned} I_A & =1.8 \mathrm{~W} / \mathrm{m}^2 \rightarrow \mathrm{III} \\ I_B & =1 \mathrm{~W} / \mathrm{m}^2 \rightarrow \mathrm{II} \\ I_C & =0.5 \mathrm{~W} / \mathrm{m}^2 \rightarrow \mathrm{IV} \end{aligned} $$ Hence, consequently option (d) is correct.

Hence, the correct answer is option: 'D'.