The wavelength of the first spectral line in the Balmer series of hydrogen atom is $6561\,\mathop {\rm{A}}\limits^ \circ $. The wavelength of the second spectral line in the Balmer series of singly ionized helium atom is

$1215\,\mathop {\rm{A}}\limits^ \circ $

$1640\,\mathop {\rm{A}}\limits^ \circ $

$2430\,\mathop {\rm{A}}\limits^ \circ $

$4687\,\mathop {\rm{A}}\limits^ \circ $

NEET UG PHYSICS MCQ: ATOMS

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The wavelength of the first spectral line in the Balmer series of hydrogen atom is

6561\,\mathop \mathrm{A}\limits^ \circ

. The wavelength of the second spectral line in the Balmer series of singly ionized helium atom is

View answer and explanation

Correct option: A

1215\,\mathop \mathrm{A}\limits^ \circ

Option A (

1215 \mathop \mathrm{A}\limits^ \circ

) is correct because the wavelength is calculated using the Rydberg formula, considering the change in Z and n values.

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50 \mathrm{~nm}