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A cobalt ( $\mathrm{Co})$ plate is placed at a distance of $1 \mathrm{~m}$ from a point source of power $1 \mathrm{~W}$. Assume a circular area of the plate of radius, $r=1 \mathrm{~A}$ is exposed to the radiation and ejects photo electrons. The light energy is considered to be spread uniformly and the work function of cobalt is $5 \mathrm{eV}$. The minimum time the target should be exposed to the light source to eject a photoelectron (Assuming no reflection losses) is
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The correct answer is:
$320 \mathrm{~s}$
Given, source power, $P=1 \mathrm{~W}$
Intensity of light source at a distance of $1 \mathrm{~m}$ is given by
$$
I=\frac{P}{4 \pi r^2}=\frac{1}{4 \pi .1^2}=\frac{1}{4 \pi} \mathrm{W} / \mathrm{m}^2
$$
Power absorbed by the circular plates of radius $1 Å$ is; $P=I \times$ Area of plate,
$$
\begin{aligned}
& =\frac{1}{4 \pi} \times \pi r^2 \quad\left[\because r=1 Å=10^{-10} \mathrm{~m}\right] \\
& P=\frac{1}{4}\left(10^{-10}\right)^2=\frac{10^{-20}}{4}
\end{aligned}
$$
Work function of cobalt
$$
\begin{aligned}
W & =5 \mathrm{eV}=5 \times 1.6 \times 10^{-19} \mathrm{~J} \\
P & =\frac{W}{t} \\
t & =\frac{W}{P} \\
\therefore \quad t & =\frac{5 \times 1.6 \times 10^{-19}}{\frac{10^{-20}}{4}}=320 \mathrm{~s}
\end{aligned}
$$
Intensity of light source at a distance of $1 \mathrm{~m}$ is given by
$$
I=\frac{P}{4 \pi r^2}=\frac{1}{4 \pi .1^2}=\frac{1}{4 \pi} \mathrm{W} / \mathrm{m}^2
$$
Power absorbed by the circular plates of radius $1 Å$ is; $P=I \times$ Area of plate,
$$
\begin{aligned}
& =\frac{1}{4 \pi} \times \pi r^2 \quad\left[\because r=1 Å=10^{-10} \mathrm{~m}\right] \\
& P=\frac{1}{4}\left(10^{-10}\right)^2=\frac{10^{-20}}{4}
\end{aligned}
$$
Work function of cobalt
$$
\begin{aligned}
W & =5 \mathrm{eV}=5 \times 1.6 \times 10^{-19} \mathrm{~J} \\
P & =\frac{W}{t} \\
t & =\frac{W}{P} \\
\therefore \quad t & =\frac{5 \times 1.6 \times 10^{-19}}{\frac{10^{-20}}{4}}=320 \mathrm{~s}
\end{aligned}
$$
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