Question

When a centimetre thick surface is illuminated with light of wavelength $\lambda$, the stopping potential is $V$. When the same surface is illuminated by light of wavelength $2\lambda$, the stopping potential is $\frac{V}{3}$. The threshold wavelength for the surface is

• A. $\frac{4\lambda }{3}$
• B. $4\lambda$
• C. $6\lambda$
• D. $\frac{8\lambda }{3}$

Answer 1

The correct option is B $4\lambda$

Given,

For $\lambda \to V2\lambda \to \frac{V}{3}$

From, the Einstein's photoelectric equation,

$E={\varphi }_0+K{E}_{max}$

$\Rightarrow eV=E-{\varphi }_0$

$\Rightarrow eV=h\nu -h{\nu }_0$

$\Rightarrow eV=\frac{hc}{\lambda }-\frac{hc}{{\lambda }_0}=hc(\frac{1}{\lambda }-\frac{1}{{\lambda }_0})$

For wavelength $\left(\lambda \right)$ ;

$eV=hc(\frac{1}{\lambda }-\frac{1}{{\lambda }_0})......\left(1\right)$

For wavelength $\left(2\lambda \right)$ ;

$\frac{eV}{3}=hc(\frac{1}{2\lambda }-\frac{1}{{\lambda }_0})\text{(or)}$

$eV=3hc(\frac{1}{2\lambda }-\frac{1}{{\lambda }_0}).......\left(2\right)$

From (1) and (2) we get,

$\frac{1}{\lambda }-\frac{1}{{\lambda }_0}=3(\frac{1}{2\lambda }-\frac{1}{{\lambda }_0})$

$\Rightarrow {\lambda }_0=4\lambda$

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