JEE PYQ: Atomic Physics Question 50
Question 50 - 2019 (10 Apr Shift 1)
In a photoelectric effect experiment the threshold wavelength of light is 380 nm. If the wavelength of incident light is 260 nm, the maximum kinetic energy of emitted electrons will be:
Given $E$ (in eV) $= \frac{1237}{\lambda (\text{in nm})}$
(1) 1.5 eV
(2) 3.0 eV
(3) 4.5 eV
(4) 15.1 eV
Show Answer
Answer: (1)
Solution
$KE_{max} = E - \phi_0$ (where $E$ = energy of incident light, $\phi_0$ = work function). $= \frac{hc}{\lambda} - \frac{hc}{\lambda_0} = 1237\left[\frac{1}{260} - \frac{1}{380}\right] = \frac{1237 \times 120}{380 \times 260} = 1.5$ eV.
BETA
