Question

If stopping potentials corresponding to wavelengths 4000 $A^0$ are 1.3V and 0.9V respectively, then the work function of the metal is

• A. 0. 3 eV
• B. 1 .3 eV
• C. 2.3 eV
• D. 5 eV

Answer 1

Answer: (C)

$e{V}_S=\frac{hc}{\lambda }-\varphi \text{ or }e{V}_S+{\varphi }_0=\frac{hc}{\lambda }$
$or\lambda =\frac{hc}{e{V}_S+{\varphi }_0}$
$\Rightarrow \frac{{\lambda }_2}{{\lambda }_1}=\frac{e{V}_{S1}+{\varphi }_0}{e{V}_{S2}+{\varphi }_0}$
$or\frac{4500}{4000}=\frac{1.3+{\varphi }_0}{0.9+{\varphi }_0}$
$or{\varphi }_0=851.-9.\times 0.9$
$Solving:{\varphi }_0=(10.4-8.1)=2.3eV$