Question

Two identical photocathodes receive the light of frequencies $f_1$ and $f_2$ respectively. If the velocities of the photo-electrons 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. $v_1-v_2={\left[\frac{2h}{m}\right(f_1-f_2\left)\right]}^{\frac{1}{2}}$
• C. $v_1^2+v_2^2=\frac{2h}{m}[f_1+f_2]$
  
• D. $v_1^2-v_2^2=\frac{2h}{m}[f_1-f_2]$
  

Answer 1

The correct option is D $v_1^2-v_2^2=\frac{2h}{m}[f_1-f_2]$


Using expression for $K.E.$ of photoelectrons
$\frac{1}{2}m{v}_1^2=h{f}_1-\varphi$ ...(1)

$\frac{1}{2}m{v}_2^2=h{f}_2-\varphi$ ...(2)
[$\varphi$ is the same for same material of photocathodes]

Subtracting equation (2) from equation (1)
$\frac{1}{2}m{v}_1^2-\frac{1}{2}m{v}_2^2=h{f}_1-h{f}_2$
$\Rightarrow v_1^2-v_2^2=\frac{2h}{m}(f_1-f_2)$