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Question

# let f2 be the frequency of second line of Lyman series and f1 be the frequency of first line of Balmer series, then the frequency of first line of Lyman series is given by,

A
f2f1
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B
f2+f1
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C
f1f2
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D
f1f2f1+f2
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Solution

## The correct option is A f2−f1For second line of Lyman series, f2=R c(112−132) ___(1) For first line of Balmer series, f1=R c(122−132) ___(2) Subtracting (2) from (1) we get, ⇒f=R c(112−122) The above equation gives the frequency of the first member of Lyman series. ⇒f=f2−f1 Hence, (A) is the correct answer. Why this question?This question is based on the spectral line of hydrogen atom.

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