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(B) The de Broglie wavelength of an electron accelerated through a potential $V$ is given by $\lambda = \frac{h}{\sqrt{2meV}}$.The wavelength of the f

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$\frac{9R^2 h^2}{2me}$

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(B) The de Broglie wavelength of an electron accelerated through a potential $V$ is given by $\lambda = \frac{h}{\sqrt{2meV}}$.The wavelength of the first line of the Lyman series for a hydrogen-like ion with atomic number $Z$ is given by $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.For $He^+$,$Z = 2$. For the first line of the Lyman series,$n_1 = 1$ and $n_2 = 2$.$\frac{1}{\lambda} = R (2)^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = 4R \left( 1 - \frac{1}{4} \right) = 4R \left( \frac{3}{4} \right) = 3R$.Equating the two wavelengths: $\frac{h}{\sqrt{2meV}} = \frac{1}{3R}$.Squaring both sides: $\frac{h^2}{2meV} = \frac{1}{9R^2}$.Solving for $V$: $V = \frac{9R^2 h^2}{2me}$.

With what potential an electron should be accelerated so that its de Broglie wavelength becomes equal to the wavelength of the first line of the Lyman series for the $He^+$ ion?

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