Question

Calculate the stopping potential $V_{S_2}$.

• A. 5.00 V
• B. 7.70 V
• C. 4.00 V
• D. 4.47 V

Answer 1

The correct option is D 4.47 V

Using $e{V}_s=\frac{hc}{\lambda }-\varphi$
where $\varphi$ is the work function of metal, $hc=1241eV$ if wavelength is put in $nm$.
Case 1: $\lambda =550nm$ $V_s=0.19V$
$\therefore$ $0.19eV=\frac{1241}{550}eV-\varphi$
Or $0.19eV=2.256eV-\varphi$
$⟹\varphi =2.07eV$
Case 2 : $\lambda =190nm$
$\therefore$ $e{V}_{s_2}=\frac{1241}{190}-\varphi$
Or $e{V}_{s_2}=6.532-2.07=4.47eV$
$⟹V_{s_2}=4.47V$