Question:Two identical photocathodes receive the light of frequencies $f_{1}$ and $f_{2}$ respectively. If the velocities of the photo-electrons coming out are $v_{1}$ and $v_{2}$ respectively, then
$\mathrm{v}_{1}^{2}-\mathrm{v}_{2}^{2}=\frac{2 \mathrm{~h}}{\mathrm{~m}}\left[\mathrm{f}_{1}-\mathrm{f}_{2}\right]$
$\mathrm{v}_{1}^{2}+\mathrm{v}_{2}^{2}=\frac{2 \mathrm{~h}}{\mathrm{~m}}\left[\mathrm{f}_{1}+\mathrm{f}_{2}\right]$
$v_{1}+v_{2}=\left[\frac{2 h}{m}\left(f_{1}+f_{2}\right)\right]^{\frac{1}{2}}$
$\mathrm{v}_{1}-\mathrm{v}_{2}=\left[\frac{2 \mathrm{~h}}{\mathrm{~m}}\left(\mathrm{f}_{1}-\mathrm{f}_{2}\right)\right]^{1 / 2}$
Correct Option: 1
Solution:(1) $\frac{1}{2} \mathrm{mv}_{1}^{2}=\mathrm{hf}_{1}-\phi$
$\frac{1}{2} \mathrm{mv}_{2}^{2}=\mathrm{hf}_{2}-\phi$
$v_{1}^{2}-v_{2}^{2}=\frac{2 h}{m}\left(f_{1}-f_{2}\right)$
