JEE PYQ: Experimental Physics Question 20
Question 20 - 2019 (09 Apr 2019 Shift 1)
The electric field of light wave is given as $\vec{E} = 10^3 \cos\left(\frac{2\pi x}{5 \times 10^{-7}} - 2\pi \times 6 \times 10^{14} t\right) \hat{x} \frac{N}{C}$. This light falls on a metal plate of work function 2 eV. The stopping potential of the photo-electrons is: (Given, E (in eV) $= \frac{12375}{\lambda (\text{in } \text{\AA})}$)
(1) 2.0 V
(2) 0.72 V
(3) 0.48 V
(4) 2.48 V
Show Answer
Answer: (3)
Solution
$f = 6 \times 10^{14}$ Hz. $\lambda = \frac{c}{f} = \frac{3 \times 10^8}{6 \times 10^{14}} = 5000$ A. $E = \frac{12375}{5000} = 2.48$ eV. $V_s = E - W = 2.48 - 2 = 0.48$ V.
BETA
