JEE PYQ: Atomic Physics Question 37
Question 37 - 2020 (07 Jan Shift 1)
A beam of electromagnetic radiation of intensity $6.4 \times 10^{-5}$ W/cm$^2$ is comprised of wavelength, $\lambda = 310$ nm. It falls normally on a metal (work function $\phi = 2$ eV) of surface area of 1 cm$^2$. If one in $10^3$ photons ejects an electron, total number of electrons ejected in 1 s is $10^x$. ($hc = 1240$ eV$\cdot$nm, $1$ eV $= 1.6 \times 10^{-19}$ J), then $x$ is ____.
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Answer: 11
Solution
Energy of photon $E = \frac{hc}{\lambda} = \frac{1240}{310} = 4$ eV $> 2$ eV $(= \phi)$, so emission will take place. $= 4 \times 1.6 \times 10^{-19} = 6.4 \times 10^{-19}$ joule. $N = \frac{6.4 \times 10^{-5} \times 1}{4 \times 6.4 \times 10^{-19}} = \frac{10^{14}}{4}$. No. of photoelectrons emitted per second $= \frac{10^{14}/4}{10^3} = 10^{11}$. Value of $x = 11$.
BETA
