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MHT CET 2026 (19 Apr, Shift 1) · Hydrogen Spectrum & Spectral Series

Hydrogen Spectrum & Spectral Series question from MHT CET 2026 (19 Apr, Shift 1)

Q7MHT CET 2026 · 19 APR · SHIFT 1Medium

Hydrogen Spectrum & Spectral Series

The shortest wavelength in the Balmer series of hydrogen atom is equal to the shortest wavelength in the Brackett series of a hydrogen like atom of atomic number 'Z'. The value of 'Z' is

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Answer: (d)

Shortest $\lambda$ = series limit: $\frac{1}{\lambda} = \frac{RZ^2}{n_1^2}$ .
H Balmer: $\lambda = \frac{4}{R}$ . Brackett ( $n_1 = 4$ ) for atomic number $Z$ : $\lambda = \frac{16}{RZ^2}$ .
$\frac{4}{R} = \frac{16}{RZ^2} \Rightarrow Z^2 = 4 \Rightarrow Z = 2$ .
So the answer is (d) $2$ .

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