Question

When a certain metallic surface is illuminated with monochromatic light of wavelength $\lambda ,$ the stopping potential $4.5\text{V}.$ When it is illuminated with light of wave length $2\lambda ,$ the stopping potential is $1.5\text{V}.$ The maximum wavelength of incident light that can release photo electrons from this surface is-

• A. $6\lambda$
• B. $4\lambda$
• C. $8\lambda$
• D. $2\lambda$

Answer 1

The correct option is B $4\lambda$

From Einstein's photo electric equation,

$K{E}_{max}=h\nu -h{\nu }_0[\because K{E}_{max}=eV]$

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

Where, $\lambda =\text{Wavelength of the incident light}{\lambda }_0=\text{Threshold wavelength}$

Let, ${\lambda }_1=\lambda ;{\lambda }_2=2\lambda V_1=4.5\text{V};V_2=1.5\text{V}$

From (1) we get,

$hc[\frac{1}{{\lambda }_1}-\frac{1}{{\lambda }_0}]=e{V}_1$

$hc\left[\frac{{\lambda }_0-\lambda }{\lambda {\lambda }_0}\right]=e{V}_1......\left(2\right)$

Similarly,

$hc[\frac{1}{{\lambda }_2}-\frac{1}{{\lambda }_0}]=e{V}_2$

$hc\left[\frac{{\lambda }_0-2\lambda }{2\lambda {\lambda }_0}\right]=e{V}_2......\left(3\right)$

From, (2) and (3) we get,

$\left[\frac{{\lambda }_0-\lambda }{\lambda {\lambda }_0}\right]\times \left[\frac{2\lambda {\lambda }_0}{{\lambda }_0-2\lambda }\right]=\frac{V_1}{V_2}=\frac{4.5}{1.5}=3$

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

$2{\lambda }_0-2\lambda =3{\lambda }_0-6\lambda$

${\lambda }_0=4\lambda$

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