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When a metal surface is exposed to certain frequency of electromagnetic radiation. The kinetic energy of electron ejected from metal surface is $0.20 \mathrm{eV}$. If its work function $\left(W_0\right)$ is $4.80 \mathrm{eV}$, the approximate frequency of radiation falling on the metal surface in $\mathrm{Hz}$ is
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Verified Answer
The correct answer is:
$1.21 \times 10^{15}$
Given, work function $\left(\omega_0\right)$ i.e. energy
for ejection $=4.80 \mathrm{eV}$ and kinetic energy of the ejected electron.
$$
(\mathrm{KE})=0.20 \mathrm{eV} \text {. }
$$
According to photoelectric current relation.
Total energy
$$
\begin{aligned}
\mathrm{Y}\left(E_T\right) & =\omega_0 t+\mathrm{KE}=4.80+0.20 \\
E_T & =5.0 \mathrm{eV}
\end{aligned}
$\begin{aligned} & E_T=5.0 \mathrm{eV} \\ & \text { Also, } \quad 1 \mathrm{eV}=1.6021 \times 10^{-19} \mathrm{~J} \\ & \therefore \quad 5 \mathrm{eV}=8.010 \times 10^{-19} \mathrm{~J} \\ & \text { or } \\ & E_T=8.010 \times 10^{-19} \mathrm{~J} \\ & \end{aligned}$
$$
$$
\begin{aligned}
& \text { and frequency }(\mathrm{v})=\frac{E_T(\text { Total energy })}{h(\text { Planks constant })} \\
& \left(\text { where, } h=6.6260 \times 10 \mathrm{~m}^2 \mathrm{~kg} / \mathrm{s}\right) \\
& \therefore \quad V=\frac{E_T}{h}=\frac{8.010 \times 10^{-19}}{6.626 \times 10^{-34}}
\end{aligned}
$$
$$
\therefore \quad V=\frac{E_T}{h}=\frac{8.010 \times 10^{-19}}{6.626 \times 10^{-34}}
$$
Frequency $(V)=1.21 \times 10^{15} \mathrm{~Hz}$.
for ejection $=4.80 \mathrm{eV}$ and kinetic energy of the ejected electron.
$$
(\mathrm{KE})=0.20 \mathrm{eV} \text {. }
$$
According to photoelectric current relation.
Total energy
$$
\begin{aligned}
\mathrm{Y}\left(E_T\right) & =\omega_0 t+\mathrm{KE}=4.80+0.20 \\
E_T & =5.0 \mathrm{eV}
\end{aligned}
$\begin{aligned} & E_T=5.0 \mathrm{eV} \\ & \text { Also, } \quad 1 \mathrm{eV}=1.6021 \times 10^{-19} \mathrm{~J} \\ & \therefore \quad 5 \mathrm{eV}=8.010 \times 10^{-19} \mathrm{~J} \\ & \text { or } \\ & E_T=8.010 \times 10^{-19} \mathrm{~J} \\ & \end{aligned}$
$$
$$
\begin{aligned}
& \text { and frequency }(\mathrm{v})=\frac{E_T(\text { Total energy })}{h(\text { Planks constant })} \\
& \left(\text { where, } h=6.6260 \times 10 \mathrm{~m}^2 \mathrm{~kg} / \mathrm{s}\right) \\
& \therefore \quad V=\frac{E_T}{h}=\frac{8.010 \times 10^{-19}}{6.626 \times 10^{-34}}
\end{aligned}
$$
$$
\therefore \quad V=\frac{E_T}{h}=\frac{8.010 \times 10^{-19}}{6.626 \times 10^{-34}}
$$
Frequency $(V)=1.21 \times 10^{15} \mathrm{~Hz}$.
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