Question
The work functions of two metals A and B are 2 eV and 4 eV respectively. Whose threshold wavelength for photoelectric effect will be lower?
✓
Answer
$\phi_0=\frac{ hc }{\lambda_0} \Rightarrow \lambda_0=\frac{ hc }{\phi_0} \Rightarrow \lambda_0 \propto \frac{1}{\phi_0}$.
For metal $B , \phi$ is higher. So threshold wavelength for it is lower.
For metal $B , \phi$ is higher. So threshold wavelength for it is lower.
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