JEE PYQ: Atomic Physics Question 8
Question 8 - 2021 (17 Mar Shift 2)
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
(1) $v_1^2 - v_2^2 = \frac{2h}{m}[f_1 - f_2]$
(2) $v_1^2 + v_2^2 = \frac{2h}{m}[f_1 + f_2]$
(3) $v_1 + v_2 = \left[\frac{2h}{m}(f_1 + f_2)\right]^{1/2}$
(4) $v_1 - v_2 = \left[\frac{2h}{m}(f_1 - f_2)\right]^{1/2}$
Show Answer
Answer: (1)
Solution
$\frac{1}{2}mv_1^2 = hf_1 - \phi$ and $\frac{1}{2}mv_2^2 = hf_2 - \phi$. Subtracting: $v_1^2 - v_2^2 = \frac{2h}{m}(f_1 - f_2)$.
BETA
