Question

In a photocell experiment stopping potentials corresponding to 4000 $\mathring{A}$ and 5000 $\mathring{A}$ are obtained as 1.5 V and 0.8 V respectively. Work function of metal is

• A. eV
• B. 2eV
• C. 2.5eV
• D. 3 eV

Answer 1

Answer: (B)

Using $\frac{hc}{e\lambda}$ = W + V i.e. $\frac{hc}{e} \begin{Bmatrix} \frac{1}{\lambda_1} - \frac{1}{\lambda_2} \end{Bmatrix}$ = V1 - V2 i.e. $\frac{hc}{e} \begin{Bmatrix} \frac{\lambda_2 - \lambda_1}{\lambda_1 \, \lambda_2} \end{Bmatrix}$ = 0.7 i.e.,
$\frac{hc}{e} \begin{Bmatrix} \frac{1 \times 10^{-7}}{4 \times 10^{-7} \times 5 \times 10^{-7}} \end{Bmatrix}$ = 0.7 i.e., $\frac{hc}{e} = 14 \times 0^7$
W = $\frac{hc}{e \lambda_1} $ -VI = $\frac{14 \times 10^{-7}}{4 \times 10^{-7}}$ - 15 = 3.5 - 1.5 = 2eV