Question

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

• 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

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

Using Einstein photoelectric equation $E=W_0+K_{max}$
$h{f}_1=W_0+\frac{1}{2}m{v}_1^2$ . . . . . .(i)
$h{f}_2=W_0+\frac{1}{2}m{v}_2^2$ . . . . . . .(ii)
$\Rightarrow h(f_1-f_2)=\frac{1}{2}m(v_1^2-v_2^2)\Rightarrow (v_1^2-v_2^2)=\frac{2h}{m}(f_1-f_2)$