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In a photoelectric effect experiment, the kinetic energy of an emitted electron is $1.986 \times 10^{-19} \mathrm{~J}$, when a radiation of frequency $1.0 \times 10^{15} \mathrm{~s}^{-1}$ hits the metal. What is the threshold frequency of the metal $\left(\right.$ in s $\left.^{-1}\right)$ ?
(Planck's constant $=6.62 \times 10^{-34} \mathrm{~J} \mathrm{~s}$ )
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(Planck's constant $=6.62 \times 10^{-34} \mathrm{~J} \mathrm{~s}$ )
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Verified Answer
The correct answer is:
$7.0 \times 10^{14}$
Given, kinetic energy of an emitted electron
$$
=1.986 \times 10^{-19} \mathrm{~J}
$$
Frequency of radiation $=1 \times 10^{15} \mathrm{~s}^{-1}$
Threshold frequency, $v_0=$ ?
From equation,
$$
\begin{aligned}
& h v-h v_0=\mathrm{KE} \\
& h\left(v-v_0\right)=\mathrm{KE}
\end{aligned}
$$
$$
\begin{aligned}
& 6.62 \times 10^{-34} \mathrm{~J}-\mathrm{s}\left(1 \times 10^{15} \mathrm{~s}^{-1}-\mathrm{v}_0\right)=1.986 \times 10^{-19} \mathrm{~J} \\
& 1 \times 10^{15} \mathrm{~s}^{-1}-\mathrm{v}_0=\frac{1.986 \times 10^{-19} \mathrm{~J}}{6.62 \times 10^{-34} \mathrm{~J}-\mathrm{s}} \\
& v_0=1 \times 10^{15} \mathrm{~s}^{-1}-\frac{1.986 \times 10^{-19} \mathrm{~J}}{6.62 \times 10^{-34} \mathrm{~J}-\mathrm{s}} \\
& v_0=1 \times 10^{15}-0.3 \times 10^{15} \mathrm{~s}^{-1} \\
& v_0=7.0 \times 10^{14} \mathrm{~s}^{-1}
\end{aligned}
$$
$$
=1.986 \times 10^{-19} \mathrm{~J}
$$
Frequency of radiation $=1 \times 10^{15} \mathrm{~s}^{-1}$
Threshold frequency, $v_0=$ ?
From equation,
$$
\begin{aligned}
& h v-h v_0=\mathrm{KE} \\
& h\left(v-v_0\right)=\mathrm{KE}
\end{aligned}
$$
$$
\begin{aligned}
& 6.62 \times 10^{-34} \mathrm{~J}-\mathrm{s}\left(1 \times 10^{15} \mathrm{~s}^{-1}-\mathrm{v}_0\right)=1.986 \times 10^{-19} \mathrm{~J} \\
& 1 \times 10^{15} \mathrm{~s}^{-1}-\mathrm{v}_0=\frac{1.986 \times 10^{-19} \mathrm{~J}}{6.62 \times 10^{-34} \mathrm{~J}-\mathrm{s}} \\
& v_0=1 \times 10^{15} \mathrm{~s}^{-1}-\frac{1.986 \times 10^{-19} \mathrm{~J}}{6.62 \times 10^{-34} \mathrm{~J}-\mathrm{s}} \\
& v_0=1 \times 10^{15}-0.3 \times 10^{15} \mathrm{~s}^{-1} \\
& v_0=7.0 \times 10^{14} \mathrm{~s}^{-1}
\end{aligned}
$$
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