JEE PYQ: Atomic Physics Question 54
Question 54 - 2019 (12 Apr Shift 1)
The stopping potential $V_0$ (in volt) as a function of frequency ($\nu$) for a sodium emitter, is shown in the figure. The work function of sodium, from the data plotted in the figure, will be:
(Given: Planck’s constant $h = 6.63 \times 10^{-34}$ Js, electron charge $e = 1.6 \times 10^{-19}$ C)
(Graph shows linear plot of $V_0$ vs $\nu (10^{14}$ Hz), with threshold at about $\nu = 4 \times 10^{14}$ Hz)
(1) 1.82 eV
(2) 1.66 eV
(3) 1.95 eV
(4) 2.12 eV
Show Answer
Answer: (2)
Solution
$f_0 = 4 \times 10^{14}$ Hz. $W_0 = hf_0 = 6.63 \times 10^{-34} \times (4 \times 10^{14})$. $= \frac{(6.63 \times 10^{-34}) \times (4 \times 10^{14})}{1.6 \times 10^{-19}} = 1.66$ eV.
BETA
