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Question: Answered & Verified by Expert
In the spectrum of hydrogen atom, the ratio of the longest wavelength in Lyman series to the longest wavelength in the Balmer series is
PhysicsAtomic PhysicsJEE Main
Options:
  • A 527
  • B 193
  • C 49
  • D 32
Solution:
1540 Upvotes Verified Answer
The correct answer is: 527
When an atom comes down from some higher energy level to the first energy level then emitted lines form of Lyman series.

               1λL = R112-1n2

where R is Rydberg's constant.

When an atom comes from higher energy level to the second level, then Balmer series are obtained.

                   1λB = R122-1n2

For maximum wavelength

n = 2 ,1λL = R1-1(2)2 = R1-14=3R4     . (i)

n=3,1λB = R1(2)2-1(3)2 = R536                 (ii)

Dividing Eq. (ii) by Eq. (i), we get

              λLλB=527

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