Question

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

• A. $0.5$V
• B. $1.0$V
• C. $2.0$V
• D. $1.5$V

Answer 1

Answer: (B)

$1.0$V

$\frac{hc}{{\lambda }_1}=\varphi +e{V}_1$ .(i)
$\frac{hc}{{\lambda }_2}=\varphi +e{V}_2$ (ii)
(i)-(ii)
$hc(\frac{1}{{\lambda }_1}-\frac{1}{{\lambda }_2})=e(V_1-V_2)$
$\Rightarrow V_1-V_2=\frac{hc}{e}\left(\frac{{\lambda }_2-{\lambda }_1}{{\lambda }_1-{\lambda }_2}\right)$
$=(1240nm-V)\frac{100nm}{300nm\times 400nm}$
$=1V$.