JEE PYQ: Atomic Physics Question 19
Question 19 - 2021 (25 Feb Shift 2)
The stopping potential for electrons emitted from a photosensitive surface illuminated by light of wavelength 491 nm is 0.710 V. When the incident wavelength is changed to a new value, the stopping potential is 1.43 V. The new wavelength is:
(1) 400 nm
(2) 382 nm
(3) 309 nm
(4) 329 nm
Show Answer
Answer: (2)
Solution
From the photoelectric effect equation: $\frac{hc}{\lambda} = \phi + eV_s$. So $eV_{s1} = \frac{hc}{\lambda_1} - \phi$ …(i) and $eV_{s2} = \frac{hc}{\lambda_2} - \phi$ …(ii). Subtracting: $(0.710 - 1.43) = 1240\left(\frac{1}{\lambda_1} - \frac{1}{\lambda_2}\right)$. Solving gives $\lambda_2 \approx 382$ nm.
BETA
