The wavelength of the first line of Lyman series for hydrogen atom is… | NEET 2011 (AIPMT, Prelims)

Q58NEET 2011 · AIPMT · PRELIMSMedium

Hydrogen Spectrum & Spectral Series

The wavelength of the first line of Lyman series for hydrogen atom is equal to that of the second line of Balmer series for a hydrogen like ion. The atomic number Z of hydrogen like ion is

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Answer: (d)

$\frac{1}{\lambda} = RZ^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$
H, Lyman first line ( $2 \to 1$ ): $\frac{1}{\lambda} = \frac{3R}{4}$
Ion, Balmer second line ( $4 \to 2$ ): $\frac{1}{\lambda} = RZ^2\cdot\frac{3}{16}$
$\frac{3Z^2}{16} = \frac{3}{4} \Rightarrow Z^2 = 4 \Rightarrow Z = 2$
So the answer is (d) 2.

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