Question

Two identical, photocathodes 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 (where the symbols have their usual meaning)

• A. $v_1^2-v_2^2=\frac{2h}{m}(f_1-f_2)$
• 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. $v_1-v_2={\left(\frac{2h}{m}\right(f_1-f_2\left)\right)}^{\frac{1}{2}}$

Answer 1

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

$hf=h{f}_0+\frac{1}{2}m{v}^{2}Here,v_1^2=\frac{2h{f}_1}{m}-\frac{2h{f}_2}{m}\Rightarrow v_1^2-v_2^2=\frac{2h}{m}(f_1-f_2)$