If ultraviolet radiation of \(6.2 \mathrm{eV}\) falls of an aluminium surface, then kinetic energy of the…
If ultraviolet radiation of \(6.2 \mathrm{eV}\) falls of an aluminium surface, then kinetic energy of the fastest emitted electron is (work-function \(=4.2 \mathrm{eV}\) )
\(3.2 \times 10^{-19} \mathrm{~J}\)
\(32 \times 10^{-21} \mathrm{~J}\)
\(7 \times 10^{-25} \mathrm{~J}\)
\(9 \times 10^{-31} \mathrm{~J}\)
Solution
Energy of ultraviolet radiation,
\(E=6.2 \mathrm{eV}\)
Work-function, \(\phi=4.2 \mathrm{eV}\)
\(\therefore\) According to Einstein's photoelectric equation, kinetic energy of the fastest emitted electron,
$\begin{aligned}
\frac{1}{2} m v_{\text{max}}^{2} & =E-\phi=6.2-4.2=2 \text{eV} \\
& =2 \times 1.6 \times 10^{-19} \text{J} \\
\Rightarrow \quad \frac{1}{2} m v_{\text{max}}^{2} & =3.2 \times 10^{-19} \text{J}
\end{aligned}$