Question

In a phtoelectric experiment, ultraviolet light of wavelength $280\text{nm}$ is used with lithium cathode, having work function $\Phi =25\text{eV}$. If the wavelength incident light is switched to $400\text{nm}$, find out the change in the stopping potential.
$(h=6.63\times {10}^{-34}\text{Js},c=3\times {10}^8{\text{ms}}^{-1}$)

• A. $0.6\text{V}$
• B. $1.1\text{V}$
• C. $1.9\text{V}$
• D. $1.3\text{V}$

Answer 1

The correct option is D $1.3\text{V}$

$h=6.63\times {10}^{-34}\text{Js}$
$c=3\times {10}^8{\text{ms}}^{-1}$
${\lambda }_1=280\text{nm}$
${\lambda }_2=400\text{nm}$
$\Phi =25\text{eV}$

According to Einstein's photoelectric equation,
$\frac{hc}{{\lambda }_1}=\Phi +e{V}_{s1}$
$\frac{hc}{{\lambda }_2}=\Phi +e{V}_{s2}$

As ${\lambda }_1<{\lambda }_2\Rightarrow V_{s1}>V_{s2}$

$\therefore hc(\frac{1}{{\lambda }_1}-\frac{1}{{\lambda }_2})=e(V_{s1}-V_{s2})$

$V_{s1}-V_{s2}=\frac{hc}{e}(\frac{1}{{\lambda }_1}-\frac{1}{{\lambda }_2})$

$=\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{1.6\times {10}^{-19}}(\frac{1}{280\times {10}^{-9}}-\frac{1}{400\times {10}^{-9}})$
$=1.3\text{V}$

Hence, $\left(A\right)$ is the correct answer.