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  4. If stopping potentials corresponding to wavelengths 4000 A0 are 1.3V and 0.9V respectively, then the work function of the metal is

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

Dual Nature of Radiation and Matter

Solution:

$e V_{S}=\frac{h c}{\lambda}-\phi \text { or } e V_{S}+\phi_{0}=\frac{h c}{\lambda} $
$ or \lambda=\frac{h c}{e V_{S}+\phi_{0}} $
$\Rightarrow \frac{\lambda_{2}}{\lambda_{1}}=\frac{e V_{S 1}+\phi_{0}}{e V_{S 2}+\phi_{0}} $
$ or \frac{4500}{4000}=\frac{1.3+\phi_{0}}{0.9+\phi_{0}} $
$ { or } \phi_{0}=851 .-9 . \times 0.9 $
$ Solving: \phi_{0}=(10.4-8.1)=2.3 e V$