Q.

For any given series of spectral lines of atomic hydrogen, letvvmax   vmin be the difference in maximum and minimum frequencies in cm–1. The ratio of ν-lyman to ν-balmer 

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a

27:5

b

4:1

c

5:4

d

9:4

answer is B.

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Detailed Solution

For Lyman series: 

ν-max=RH 112-12=RH ν-mini.=RH 112-122=3RH4 for lyman series, ν-lyman=RH4 similarly,  ν-balmer=RH9

 

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