Q140 / 180 · COMEDK 2024 (Memory Based) · Physics
When a beam of 10.6 eV photons of intensity $2.0 \mathrm{~W} / \mathrm{m}^2$ falls on a platinum surface of area $1.0 \times 10^{-4} \mathrm{~m}^2$ and work function 5.6 eV . $0.53 \%$ of the incident photons eject photoelectrons. The number of photoelectrons emitted per second is
- Correct answerA$6.25 \times 10^{11}$
- B$4.25 \times 10^{10}$
- C$6.25 \times 10^{10}$
- D$4.25 \times 10^{11}$
Solution
Energy of incident photons,
$\begin{aligned}<br/>E_1 & =10.6 \mathrm{eV}=10.6 \times 1.6 \times 10^{-19} \mathrm{~J} \\<br/>& =16.96 \times 10^{-19} \mathrm{~J}<br/>\end{aligned}$
Energy incident per unit area per unit time (intensity) $=2 \mathrm{~J}$
$\therefore$ Number of photons incident on unit area in unit time
$\begin{aligned}<br/>& =\frac{2}{16.96 \times 10^{-19}} \\<br/>& =1.18 \times 10^{18}<br/>\end{aligned}$
Therefore, number of photons incident per unit time on given area.
$\begin{aligned}<br/>\left(1.0 \times 10^{-4} \mathrm{~m}^2\right) & \\<br/>& =\left(1.18 \times 10^{18}\right)\left(1.0 \times 10^{-4}\right) \\<br/>& =1.18 \times 10^{14}<br/>\end{aligned}$
But only $0.53 \%$ of incident, photons emit photoelectrons.
$\therefore$ Number of photoelectrons emitted per second,
$\begin{aligned}<br/>& n=\left(\frac{0.53}{100}\right)\left(1.18 \times 10^{14}\right) \\<br/>& n=6.25 \times 10^{11}<br/>\end{aligned}$
$\begin{aligned}<br/>E_1 & =10.6 \mathrm{eV}=10.6 \times 1.6 \times 10^{-19} \mathrm{~J} \\<br/>& =16.96 \times 10^{-19} \mathrm{~J}<br/>\end{aligned}$
Energy incident per unit area per unit time (intensity) $=2 \mathrm{~J}$
$\therefore$ Number of photons incident on unit area in unit time
$\begin{aligned}<br/>& =\frac{2}{16.96 \times 10^{-19}} \\<br/>& =1.18 \times 10^{18}<br/>\end{aligned}$
Therefore, number of photons incident per unit time on given area.
$\begin{aligned}<br/>\left(1.0 \times 10^{-4} \mathrm{~m}^2\right) & \\<br/>& =\left(1.18 \times 10^{18}\right)\left(1.0 \times 10^{-4}\right) \\<br/>& =1.18 \times 10^{14}<br/>\end{aligned}$
But only $0.53 \%$ of incident, photons emit photoelectrons.
$\therefore$ Number of photoelectrons emitted per second,
$\begin{aligned}<br/>& n=\left(\frac{0.53}{100}\right)\left(1.18 \times 10^{14}\right) \\<br/>& n=6.25 \times 10^{11}<br/>\end{aligned}$
Hence, the correct answer is option: 'A'.