Q. Two identical, photocathodes receive light of frequencies $f_{1}\ and\ f_{2}$. If the velocities of the photoelectrons (of mass m) coming out are respectively $v_{1}$ and $v_{2}$, then
A) $v_{1}^{2} - v_{2}^{2} = \frac{2h}{m}\ (f_{1} - f_{2})$
B) $v_{1} + v_{2} = \left\lbrack \frac{2h}{m}\ \left( f_{1} + f_{2} \right) \right\rbrack^{1/2}$
C) $v_{1}^{2} + v_{2}^{2} = \frac{2h}{m}\ (f_{1} + f_{2})$
D) $v_{1} - v_{2} = \left\lbrack \frac{2h}{m}\ \left( f_{1} - f_{2} \right) \right\rbrack^{1/2}$
Explanation
(a) $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ hf = hf_{0} + \frac{1}{2}mv^{2}$
Hence, $v_{1}^{2} = \frac{2hf_{1}}{m} - \frac{2hf_{0}}{m}$
$v_{2}^{2} = \frac{2hf_{2}}{m} - \frac{2hf_{0}}{m}$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ v_{1}^{2} - v_{2}^{2} = \frac{2h}{m}\ (f_{1} - f_{2})$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.