Question

The stopping potentials are $V_1$ and $V_2$ with incident lights of wavelength ${\lambda }_1$ and ${\lambda }_2$ respectively. Then $V_1$ -$V_2=$

• A. $\frac{hc}{e}\frac{{\lambda }_1{\lambda }_2}{{\lambda }_1-{\lambda }_2}$
• B. $\frac{hc}{e}[\frac{1}{{\lambda }_1}-\frac{1}{{\lambda }_2}]$
• C. $\frac{he}{c}[\frac{1}{{\lambda }_1}-\frac{1}{{\lambda }_2}]$
• D. $\frac{he}{c{\lambda }_1{\lambda }_2}({\lambda }_1-{\lambda }_2)$

Answer 1

The correct option is B $\frac{hc}{e}[\frac{1}{{\lambda }_1}-\frac{1}{{\lambda }_2}]$

Stopping potential $=(\frac{hc}{\lambda }-\varphi )/e$

$V_1=(\frac{hc}{{\lambda }_1}-\varphi )/e$

$V_2=(\frac{hc}{{\lambda }_2}-\varphi )/e$

$V_1-V_2=(\frac{hc}{{\lambda }_1}-\varphi )/e-(\frac{hc}{{\lambda }_2}-\varphi )/e$

$=\frac{hc}{{\lambda }_{1}e}-\frac{\varphi }{e}-\frac{hc}{{\lambda }_{2}e}+\frac{\varphi }{e}$

$=\frac{hc}{e}(\frac{1}{{\lambda }_1}-\frac{1}{{\lambda }_2})$

So, the answer is option (B).