JEE MAIN - Physics (2021 - 17th March Evening Shift - No. 6)
Two identical photocathodes receive the light of frequencies f1 and f2 respectively. If the velocities of the photo-electrons coming out are v1 and v2 respectively, then
$${v_1} - {v_2} = {\left[ {{{2h} \over m}({f_1} - {f_2})} \right]^{{1 \over 2}}}$$
$$v_1^2 + v_2^2 = {{2h} \over m}[{f_1} + {f_2}]$$
$${v_1} + {v_2} = {\left[ {{{2h} \over m}({f_1} + {f_2})} \right]^{{1 \over 2}}}$$
$$v_1^2 - v_2^2 = {{2h} \over m}[{f_1} - {f_2}]$$
Explanation
$${1 \over 2}mv_1^2 = h{f_1} - \phi $$ ___________(1)
$${1 \over 2}mv_2^2 = h{f_2} - \phi $$ ___________(2)
Subtracting equation (1) by equation (2)
$${1 \over 2}mv_1^2 - {1 \over 2}mv_2^2 = h{f_1} - h{f_2}$$
$$v_1^2 - v_2^2 = {{2h} \over m}({f_1} - {f_2})$$
$${1 \over 2}mv_2^2 = h{f_2} - \phi $$ ___________(2)
Subtracting equation (1) by equation (2)
$${1 \over 2}mv_1^2 - {1 \over 2}mv_2^2 = h{f_1} - h{f_2}$$
$$v_1^2 - v_2^2 = {{2h} \over m}({f_1} - {f_2})$$
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