Question

When photons of wavelength ${\lambda }_1$ are incident on a metal plate, the corresponding stopping potential is found to be $V$. When photons of wavelength ${\lambda }_2$ are used, the corresponding stopping potential was thrice the above value. If light of wavelength ${\lambda }_3$ is used, the stopping potential for wavelength ${\lambda }_3$ in terms of ${\lambda }_1$ and ${\lambda }_2$ will be

• A. $\frac{hc}{e}\left(\frac{1}{{\lambda }_3}+\frac{1}{2{\lambda }_2}-\frac{1}{{\lambda }_1}\right]$
• B. $\frac{hc}{e}\left(\frac{1}{{\lambda }_3}+\frac{1}{2{\lambda }_2}-\frac{3}{2{\lambda }_1}\right]$
• C. $\frac{hc}{e}\left(\frac{1}{{\lambda }_3}+\frac{1}{{\lambda }_2}-\frac{1}{{\lambda }_1}\right]$
• D. $\frac{hc}{e}\left(\frac{1}{{\lambda }_3}-\frac{1}{{\lambda }_2}-\frac{1}{{\lambda }_1}\right]$

Answer 1

The correct option is B $\frac{hc}{e}\left(\frac{1}{{\lambda }_3}+\frac{1}{2{\lambda }_2}-\frac{3}{2{\lambda }_1}\right]$

${\mathrm{KE}}_{{\lambda }_1}=\frac{hc}{{\lambda }_1}-\psi =e\Delta V$
${\mathrm{KE}}_{{\lambda }_2}=\frac{hc}{{\lambda }_2}-\psi =3e\Delta V$
$\Rightarrow 3\left(\frac{hc}{{\lambda }_1}-\psi \right)=\frac{hc}{{\lambda }_2}-\psi \Rightarrow \psi =hc\left(\frac{3}{2{\lambda }_1}-\frac{1}{2{\lambda }_2}\right)$
$\Rightarrow {\mathrm{KE}}_{{\lambda }_3}=\frac{hc}{{\lambda }_3}-hc\left(\frac{3}{2{\lambda }_1}-\frac{1}{2{\lambda }_2}\right]=hc[\frac{1}{{\lambda }_3}+\frac{1}{2{\lambda }_2}-\frac{3}{2{\lambda }_1}]$
$e\Delta V=hc\left(\frac{1}{{\lambda }_3}+\frac{1}{2{\lambda }_2}-\frac{3}{2{\lambda }_1}\right]\Delta V=\frac{hc}{e}\left(\frac{1}{{\lambda }_3}+\frac{1}{2{\lambda }_2}-\frac{3}{2{\lambda }_1}\right]$