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IIT-JEE 2011 · Paper 1, Physics question 27

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

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Answer and worked solution

Official answer: A

Hydrogen Hα. 16561=R(14−19)=5R36\frac1{6561}=R\left(\frac14-\frac19\right)=\frac{5R}{36}, so 1R=911.25\frac1R=911.25 Å.

He⁺, second Balmer line (4 → 2): 1λ=4R(14−116)=3R4\frac1\lambda=4R\left(\frac14-\frac1{16}\right)=\frac{3R}4, so λ=43×911.25=1215\lambda=\frac43\times911.25=1215 Å.

(A)

Topics:Bohr model and hydrogen spectrum
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