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Question
If the wavelength of first member of Balmer series of hydrogen spectrum is 6564˙A, the wavelength of second member of Balmer series will be:
Options:
A .  1215˙A
B .  4862˙A
C .  6050˙A
D .  5892˙A
Answer: Option B
:
B
1λ1[122132]
1λ1536 . . . .(1)
1λ2[14116]
1λ2[316] . . . .(2)
λ2λ1=5×1636×3=2027
λ2=2027×6564=4862˙A

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