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Atoms and Nuclei

Section: Physics

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Question : 38 of 51

Marks: +1, -0

The wavelength of the first spectral line in the Balmer series of hydrogen atom is 6561 Å. The wavelength of the second spectral line in the Balmer series of singly-ionized helium atom is

[JEE Adv 2011 P1]

1215 Å
1640 Å
2430 Å
4687 Å

Solution:

For hydrogen-like atom, the wavelength emitted in Balmer series is given by

\frac{1}{\lambda}

=

Z^2 R \left( \frac{1}{2^2} - \frac{1}{n^2} \right)

The given wavelength of first line of hydrogen is 6561 Å, that is,

\frac{1}{6561 \times 10^{-10} \mathrm{m}}

=

R \left( \frac{1}{4} - \frac{1}{9} \right)

⇒ R =

\left( \frac{36}{5 \times 6561 \times 10^{-10}} \right) \mathrm{m}^{-1}

Therefore, for singly ionized helium, we get

\frac{1}{\lambda}

=

2^2 R \left( \frac{1}{2^2} - \frac{1}{4^2} \right)

=

R \left( 1 - \frac{1}{4} \right)

=

\frac{36 \times 3}{4 \times 5 \times 6561 \times 10^{-10} \mathrm{m}}

⇒ λ = 1215 Å

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