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Two identical photocathodes receive the light of frequencies $f_{1} a n d f_{2}$ respectively. If the velocities of the photo-electrons coming out are $v_{1} a n d v_{2}$ respectively, then

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$v_{1} - v_{2} = {[\frac{2 h}{m} (f_{1} - f_{2})]}^{\frac{1}{2}}$

$v_{1} + v_{2} = {[\frac{2 h}{m} (f_{1} + f_{2})]}^{\frac{1}{2}}$

$v_{1}^{2} + v_{2}^{2} = \frac{2 h}{m} [f_{1} + f_{2}]$

$v_{1}^{2} - v_{2}^{2} = \frac{2 h}{m} [f_{1} - f_{2}]$

answer is B.

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Detailed Solution

$E = K E + w_{0}$

$\Rightarrow K E = E - w_{0}$

$C a s e 1 : \frac{1}{2} m v_{1}^{2} = h f_{1} - w_{0}$

$C a s e 2 : \frac{1}{2} m v_{2}^{2} = h f_{2} - w_{0}$

$Subtracting, v_{1}^{2} - v_{2}^{2} = \frac{2 h}{m} (f_{1} - f_{2})$

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