JEE PYQ: Atomic Physics Question 62
Question 62 - 2019 (11 Jan Shift 2)
In a photoelectric experiment, the wavelength of the light incident on a metal is changed from 300 nm to 400 nm. The decrease in the stopping potential is close to:
$\left(\frac{hc}{e} = 1240 \text{ nm-V}\right)$
(1) 0.5 V
(2) 1.5 V
(3) 1.0 V
(4) 2.0 V
Show Answer
Answer: (3)
Solution
$\frac{hc}{\lambda_1} = \phi + eV_1$ …(i). $\frac{hc}{\lambda_2} = \phi + eV_2$ …(ii). Subtracting: $hc\left(\frac{1}{\lambda_1} - \frac{1}{\lambda_2}\right) = e(V_1 - V_2)$. $V_1 - V_2 = \frac{hc}{e}\left(\frac{\lambda_2 - \lambda_1}{\lambda_1 \lambda_2}\right) = (1240 \text{ nm-V})\left(\frac{100 \text{ nm}}{300 \text{ nm} \times 400 \text{ nm}}\right) = 1.03$ V $\approx 1$ V.
BETA
