Solve this following


Two stream of photons, possessing energies equal to twice and ten times the work function of metal are incident on the metal surface successively. The value of ratio of maximum velocities of the photoelectrons emitted in the two respective cases is $x: y$. The value of $x$ is


$\mathrm{KE}_{\max }=\mathrm{h} v-\phi$

$\frac{1}{2} m v^{2}=h v-\phi$

$v=\sqrt{\frac{2(h v-\phi)}{m}}$

Given $h v_{1}=2 \phi$

$h v_{2}=10 \phi$

$\therefore \frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=\sqrt{\frac{\mathrm{hv} \mathrm{v}_{1}-\phi}{\mathrm{hv} \mathrm{v}_{2}-\phi}}$

$\frac{\mathrm{v}_{1}}{\mathrm{v}_{2}}=\sqrt{\frac{2 \phi-\phi}{10 \phi-\phi}}=\frac{1}{3}$


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