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Q.

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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a

2430 Å

b

1215 Å

c

4687 Å

d

1640 Å

answer is A.

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Detailed Solution

First line of Balmer of Hydrogen 
 1λ1=R(1)2[122−132] where λ1=6561
Second line of Balmer of singly ionized Helium
 1λ2=R(2)2[122−142] 
Dividing   λ2= λ1x527=6561x527=1215A0
 

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