Two identical photocathodes receive the light

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

$\mathrm{v}_{1}$ and $\mathrm{v}_{2}$ respectively, then

  1. (1) $\mathrm{v}_{1}^{2}-\mathrm{v}_{2}^{2}=\frac{2 \mathrm{~h}}{\mathrm{~m}}\left[\mathrm{f}_{1}-\mathrm{f}_{2}\right]$

  2. (2) $\mathrm{v}_{1}^{2}+\mathrm{v}_{2}^{2}=\frac{2 \mathrm{~h}}{\mathrm{~m}}\left[\mathrm{f}_{1}+\mathrm{f}_{2}\right]$

  3. (3) $\mathrm{v}_{1}+\mathrm{v}_{2}=\left[\frac{2 \mathrm{~h}}{\mathrm{~m}}\left(\mathrm{f}_{1}+\mathrm{f}_{2}\right)\right]^{-\frac{1}{2}}$

  4. (4) $\mathrm{v}_{1}-\mathrm{v}_{2}=\left[\frac{2 \mathrm{~h}}{\mathrm{~m}}\left(\mathrm{f}_{1}-\mathrm{f}_{2}\right)\right]^{1 / 2}$

Correct Option: 1

Solution:

(1)

$\frac{1}{2} \mathrm{mv}_{1}^{2}=\mathrm{hf}_{1}-\phi$

$\frac{1}{2} \mathrm{mv}_{2}^{2}=\mathrm{hf}_{2}-\phi$

$\mathrm{v}_{1}^{2}-\mathrm{v}_{2}^{2}=\frac{2 \mathrm{~h}}{\mathrm{~m}}\left(\mathrm{f}_{1}-\mathrm{f}_{2}\right)$

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