MCQ
The stopping potential doubles when the frequency of the incident light changes from $v$ to $\frac{3v}{2}$ . Then the work function of the metal must be
- ✓$\frac{hv}{2}$
- B$hv$
- C$2hv$
- Dnone of the above
✓
Answer
Correct option: A.
$\frac{hv}{2}$
a
$\mathrm{hv}=\phi+\mathrm{V}_{\mathrm{s}}$ and $\mathrm{h} \frac{3}{2} \mathrm{v}=\phi+2 \mathrm{V}_{\mathrm{s}}$
$\mathrm{hv}=\phi+\mathrm{V}_{\mathrm{s}}$ and $\mathrm{h} \frac{3}{2} \mathrm{v}=\phi+2 \mathrm{V}_{\mathrm{s}}$
$\Rightarrow \phi=\frac{\mathrm{hv}}{2}$
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