What element has a hydrogen-like spectrum whose lines have wavelengths four times shorter than those of atomic hydrogen?

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Solution

$\frac{1}{λ} = R_{H} \times z^{2} [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}]$ where $z =$ atomic number.
For hydrogen, $z = 1 ⟹ \frac{1}{λ} = R_{H} [\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}]$
Since, the wavelength is $f o u r$ times shorter than that for hydrogen, $z^{2} = 4 z = 2$ .
This corresponding to helium atom $(H e^{+})$ .

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What element has a hydrogen-like spectrum whose lines have wavelengths four times shorter than those of atomic hydrogen?

1λ=RH×z2[1n21−1n22] where z= atomic number.For hydrogen, z=1⟹1λ=RH[1n21−1n22]Since, the wavelength is four times shorter than that for hydrogen, z2=4z=2.This corresponding to helium atom (He+).