Question

In a photoelectric experiment, the wavelength of the light incident on a metal is changed from $300\text{nm}$ to $400\text{nm}$. The decrease in the stopping potential is close to :
$(\frac{hc}{e}=1240\text{nm}-V)$

• A. $2.0\text{V}$
• B. $1.5\text{V}$
• C. $0.5\text{V}$
• D. $1.0\text{V}$

Answer 1

The correct option is D $1.0\text{V}$

Let $\varphi =$ work function of the metal,
$\frac{hc}{{\lambda }_1}=\varphi +e{V}_1....\left(i\right)$
$\frac{hc}{{\lambda }_2}=\varphi +e{V}_2....\left(ii\right)$

$\text{Sutracting (ii) from (i) we get}$

$hc(\frac{1}{{\lambda }_1}-\frac{1}{{\lambda }_2})=e(V_1-V_2)$
$\Rightarrow V_1-V_2=\frac{hc}{e}\left(\frac{{\lambda }_2-{\lambda }_1}{{\lambda }_1.{\lambda }_2}\right)\left[\begin{array}{c}{\lambda }_1=300\text{nm}\\ {\lambda }_2=400\text{nm}\\ \frac{hc}{e}=1240\text{nm}\end{array}\right]$

$=\left(1240\right)\left(\frac{100}{300\times 400}\right)$

$=1.03\approx 1\text{V}$

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