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The work function of a certain metal is $3.31 \times 10^{-19}$ J. Then, the maximum kinetic energy of photoelectrons emitted by incident radiation of wavelength $5000 Å$ is (Given, $\mathrm{h}=6.62 \times 10^{-34} \mathrm{~J}$ $\left.\mathrm{s}, \mathrm{c}=3 \times 10^{8} \mathrm{~ms}^{-1}, e=1.6 \times 10^{-19} \mathrm{C}\right)$
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The correct answer is:
$0.41 \mathrm{eV}$
Work function $W_{0}=3.31 \times 10^{-19} \mathrm{~J}$ Wavelength of incident radiation $\lambda=5000 \times 10^{-10} \mathrm{~m}$
According to Einstein's photoelectric equation $E=W_{0}+K E$
$\begin{aligned} \Rightarrow \frac{h c}{\lambda} &=3.31 \times 10^{-19}+\mathrm{KE} \\ \mathrm{KE} &=-3.31 \times 10^{-19}+\frac{6.62 \times 10^{-34} \times 3 \times 10^{8}}{5000 \times 10^{-10}} \\ &=-3.31 \times 10^{-19}+\frac{6.62 \times 3}{5} \times 10^{-19} \\ &=(-3.31 \times 1.324 \times 3) \times 10^{-19} \end{aligned}$
$\begin{aligned}
&=(3.972-3.31) \times 10^{-19} \\
&=0.662 \times 10^{-19} \mathrm{~J} \\
\text { or } E &=\frac{0.662 \times 10^{-19}}{1.6 \times 10^{-19}}=0.41 \mathrm{eV}
\end{aligned}$
According to Einstein's photoelectric equation $E=W_{0}+K E$
$\begin{aligned} \Rightarrow \frac{h c}{\lambda} &=3.31 \times 10^{-19}+\mathrm{KE} \\ \mathrm{KE} &=-3.31 \times 10^{-19}+\frac{6.62 \times 10^{-34} \times 3 \times 10^{8}}{5000 \times 10^{-10}} \\ &=-3.31 \times 10^{-19}+\frac{6.62 \times 3}{5} \times 10^{-19} \\ &=(-3.31 \times 1.324 \times 3) \times 10^{-19} \end{aligned}$
$\begin{aligned}
&=(3.972-3.31) \times 10^{-19} \\
&=0.662 \times 10^{-19} \mathrm{~J} \\
\text { or } E &=\frac{0.662 \times 10^{-19}}{1.6 \times 10^{-19}}=0.41 \mathrm{eV}
\end{aligned}$
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