Question
Two identical photocathodes receive light of frequencies $${f_1}$$ and $${f_2}.$$ If the velocites of the photo electrons (of mass $$m$$) coming out are respectively $${v_1}$$ and $${v_2},$$ then
A.
$$v_1^2 - v_2^2 = \frac{{2h}}{m}\left( {{f_1} - {f_2}} \right)$$
B.
$${v_1} + {v_2} = {\left[ {\frac{{2h}}{m}\left( {{f_1} + {f_2}} \right)} \right]^{\frac{1}{2}}}$$
C.
$$v_1^2 + v_2^2 = \frac{{2h}}{m}\left( {{f_1} + {f_2}} \right)$$
D.
$${v_1} - {v_2} = {\left[ {\frac{{2h}}{m}\left( {{f_1} - {f_2}} \right)} \right]^{\frac{1}{2}}}$$
Answer :
$$v_1^2 - v_2^2 = \frac{{2h}}{m}\left( {{f_1} - {f_2}} \right)$$
Solution :
For one photocathode
$$h{f_1} - W = \frac{1}{2}mv_1^2\,......\left( {\text{i}} \right)$$
For another photo cathode
$$h{f_2} - W = \frac{1}{2}mv_2^2\,......\left( {{\text{ii}}} \right)$$
Subtracting (ii) from (i) we get
$$\eqalign{
& \left( {h{f_1} - W} \right) - \left( {h{f_2} - W} \right) = \frac{1}{2}mv_1^2 - \frac{1}{2}mv_2^2 \cr
& \therefore h\left( {{f_1} - {f_2}} \right) = \frac{m}{2}\left( {v_1^2 - v_2^2} \right) \cr
& \therefore v_1^2 - v_2^2 = \frac{{2h}}{m}\left( {{f_1} - {f_2}} \right) \cr} $$