Question
With what potential an electron should be accelerated so that its de Broglie wavelength becomes equal to the wavelength of first line of lymen series for $He^+$ ion ?
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Answer
$\frac{\mathrm{h}}{\sqrt{2 \mathrm{meV}}}=\frac{1}{\mathrm{R}(2)^{2}\left(\frac{1}{1^{2}}-\frac{1}{2^{2}}\right)}$
$\Rightarrow \frac{\mathrm{h}}{\sqrt{2 \mathrm{meV}}}=\frac{1}{4 \mathrm{R}\left(\frac{3}{4}\right)}=\frac{1}{3 \mathrm{R}}$
$\frac{\mathrm{h}^{2}}{2 \mathrm{meV}}=\frac{1}{9 \mathrm{R}^{2}} \Rightarrow \mathrm{V}=\frac{9 \mathrm{R}^{2} \mathrm{h}^{2}}{2 \mathrm{me}}$
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