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. $\frac{hc}{e}\left\{\begin{array}{c}\frac{1}{{\lambda }_1}-\frac{1}{{\lambda }_2}\end{array}\right\}$ = V1 - V2 i.e. $\frac{hc}{e}\left\{\begin{array}{c}\frac{{\lambda }_2-{\lambda }_1}{{\lambda }_1{\lambda }_2}\end{array}\right\}$ = 0.7 i.e.,
$\frac{hc}{e}\left\{\begin{array}{c}\frac{1\times 10^{-7}}{4\times 10^{-7}\times 5\times 10^{-7}}\end{array}\right\}$ = 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