Question

The stopping potential for electrons emitted from a photosensitive surface illuminated by light of wavelength $491\text{nm}$ is $0.710\text{V}$. When the incident wavelength is changed to a new value, the stopping potential is $1.43\text{V}$. The new wavelength is:

• A. $400\text{nm}$
• B. $382\text{nm}$
• C. $309\text{nm}$
• D. $329\text{nm}$

Answer 1

The correct option is B $382\text{nm}$

From the photoelectric effect equation,
$\frac{hc}{\lambda }=\varphi +e{V}_s$
where $e{V}_s$ is the stopping potential and $\varphi$ is the work function of the metal.

So, $e{V}_{s_1}=\frac{hc}{{\lambda }_1}-\varphi$ ..............$\left(i\right)$

$e{V}_{s_2}=\frac{hc}{{\lambda }_2}-\varphi$ ..............$\left(ii\right)$

Subtract equation $\left(i\right)$ from equation $\left(ii\right)$,
$e{V}_{s_1}-e{V}_{s_2}=\frac{hc}{{\lambda }_1}-\frac{hc}{{\lambda }_2}$

$V_{s_1}-V_{s_2}=\frac{hc}{e}(\frac{1}{{\lambda }_1}-\frac{1}{{\lambda }_2})$

$0.710-1.43=1240(\frac{1}{491}-\frac{1}{{\lambda }_2})$
because $hc=1240\text{eV-nm}$

On solving this, we get,
${\lambda }_2\approx 382\text{nm}$