JEE MAIN - Physics (2003 - No. 5)
Two identical photo-cathodes receive light of frequencies $${f_1}$$ and $${f_2}$$. If the velocities of the photo electrons (of mass $$m$$ ) coming out are respectively $${v_1}$$ and $${v_2},$$ then
$$v_1^2 - v_2^2 = {{2h} \over m}\left( {{f_1} - {f_2}} \right)$$
$${v_1} + {v_2} = {\left[ {{{2h} \over m}\left( {{f_1} + {f_2}} \right)} \right]^{1/2}}$$
$$v_1^2 + v_2^2 = {{2h} \over m}\left( {{f_1} + {f_2}} \right)$$
$${v_1} - {v_2} = {\left[ {{{2h} \over m}\left( {{f_1} - {f_2}} \right)} \right]^{1/2}}$$
Explanation
For one photo cathode
$$h{f_1} - W = {1 \over 2}mv_1^2\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$
For another photo cathode
$$h{f_2} - W = {1 \over 2}mv_2^2\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$
Subtracting $$(ii)$$ from $$(i)$$ we get
$$\left( {h{f_1} - W} \right) - \left( {h{f_2} - W} \right) = {1 \over 2}mv_1^2 - {1 \over 2}mv_2^2$$
$$\therefore$$ $$h\left( {{f_1} - {f_2}} \right) = {m \over 2}\left( {v_1^2 - v_2^2} \right)$$
$$\therefore$$ $$v_1^2 - v_2^2 = {{2h} \over m}\left( {{f_1} - {f_2}} \right)$$
$$h{f_1} - W = {1 \over 2}mv_1^2\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$
For another photo cathode
$$h{f_2} - W = {1 \over 2}mv_2^2\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$
Subtracting $$(ii)$$ from $$(i)$$ we get
$$\left( {h{f_1} - W} \right) - \left( {h{f_2} - W} \right) = {1 \over 2}mv_1^2 - {1 \over 2}mv_2^2$$
$$\therefore$$ $$h\left( {{f_1} - {f_2}} \right) = {m \over 2}\left( {v_1^2 - v_2^2} \right)$$
$$\therefore$$ $$v_1^2 - v_2^2 = {{2h} \over m}\left( {{f_1} - {f_2}} \right)$$
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