Question

For photoelectric effect in a metal, the graph of the stopping potential $V_0\text{(in V)}$ versus frequency $\nu \text{(in Hz)}$ of the incident radiation is shown. The work function of the metal ( $\text{(in eV)}$ is-

• A. $18.5$
• B. $16.5$
• C. $14.5$
• D. $12.5$

Answer 1

The correct option is B $16.5$

From the graph we get,

The threshold frequency ${\nu }_0=4\times {10}^{15}\text{Hz}$

Work function,

$\varphi =h{\nu }_0=6.6\times {10}^{-34}\times 4\times {10}^{15}\text{J}$

$=\frac{6.6\times {10}^{-34}\times 4\times {10}^{15}}{1.6\times {10}^{-19}}\text{eV}$

$\varphi =16.5\text{eV}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(B\right)$ is the correct answer.