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