Question

Two identical photocathodes receive light of frequencies $f_1$ and $f_2$. If the velocities of the photoelectrons (of mass m) coming out are $v_1$ and $v_2$, respectively, then

• A. $v_1-v_2={\left[\frac{2h}{m}\right(f_1-f_2\left)\right]}^{\frac{1}{2}}$
• B. $v_1^2-v_2^2=\frac{2h}{m}(f_1-f_2)$
• C. $v_1+v_2={\left[\frac{2h}{m}\right(f_1-f_2\left)\right]}^{\frac{1}{2}}$
• D. $v_1^2+v_2^2=\frac{2h}{m}(f_1-f_2)$

Answer 1

The correct option is B $v_1^2-v_2^2=\frac{2h}{m}(f_1-f_2)$

Let work function be $X$ and $K_1$ and $K_2$ be kinetic energies for $f_1$ and $f_2$ respectively when frequency is $f_1$
$K_1=h{f}_1-X$ $\to 1$
$K_2=h{f}_2-X$ $\to 2$
$eq1-eq2$ =>
$k_1-K_2=h{f}_1-h{f}_2$

but $K=\frac{m{V}^2}{2}$

$\frac{m(v_1^2-v_2^2)}{2}=h(f_1-f_2)$

$v_1^2-v_2^2=\frac{2h(f_1-f_2)}{m}$