JEE PYQ: Atomic Physics Question 30
Question 30 - 2020 (04 Sep Shift 1)
Given figure shows few data points in a photo electric effect experiment for a certain metal. The minimum energy for ejection of electron from its surface is: (Planck’s constant $h = 6.62 \times 10^{-34}$ J.s)
(Graph shows $V_{stop}$ vs $f(10^{14}$ Hz) with points at A$(5, 0)$, B$(5.5, 0)$ and C$(6, V)$)
(1) 2.27 eV
(2) 2.59 eV
(3) 1.93 eV
(4) 2.10 eV
Show Answer
Answer: (1)
Solution
Work function $\phi = hV$ joule or $\phi = \frac{hf}{e}$ eV (for $V = 0$). $\phi = \frac{6.62 \times 10^{-34} \times 5.5 \times 10^{14}}{1.6 \times 10^{-19}}$ eV $= 2.27$ eV.
BETA
