JEE PYQ: Experimental Physics Question 26
Question 26 - 2019 (10 Jan 2019 Shift 2)
A metal plate of area $1 \times 10^{-4}$ m$^2$ is illuminated by a radiation of intensity 16 mW/m$^2$. The work function of the metal is 5 eV. The energy of the incident photons is 10 eV and only 10% of it produces photo electrons. The number of emitted photo electrons per second and their maximum energy, respectively, will be: [$1$ eV $= 1.6 \times 10^{-19}$ J]
(1) $10^{14}$ and 10 eV
(2) $10^{12}$ and 5 eV
(3) $10^{11}$ and 5 eV
(4) $10^{10}$ and 5 eV
Show Answer
Answer: (3)
Solution
$\frac{n}{t} = \frac{I \times A}{E} = \frac{16 \times 10^{-3} \times 10^{-4}}{10 \times 1.6 \times 10^{-19}} = 10^{12}$. Effective photoelectrons $= 10^{12} \times 0.1 = 10^{11}$. Max KE $= 10 - 5 = 5$ eV.
BETA
