Atomic Structure Question 474

Question: Last line of Lyman series for H-atom has wavelength λ1Ao,,

2nd line of Balmer series has wavelength λ2Ao, then

Options:

A) 16λ1=9λ2

B) 16λ2=9λ1

C) 4λ1=1λ2

D) 16λ1=3λ2

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Answer:

Correct Answer: B

Solution:

  • 1λ1=R(1)2(11212) and 1λ2=R(1)2(122142)

λ1=1R and λ2=163R

16λ2=3λ1



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