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Question

# The atomic spectrum of Li2+ ion arises due to the transition of an electron from n2 to n1. If n1+n2=4 and n2−n1=2, find the wavelength of 3rd line of this series in Li2+ ion?

A
1.08 nm
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B
10.8 nm
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C
108 nm
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D
1080 nm
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Solution

## The correct option is D 1080 nmAccording to the given data n1+n2=4....(i) n2−n1=2....(ii) Adding equation (i) and (ii) we get 2n2=6 ∴n2=3 From equation (i), n1+3=4 ∴ n1=1 Therefore, the series is Lyman series. The 3rd line of this series will have a transition from n2=4→n1=1 We know, 1λ=RZ2[1n21−1n22] =1.097×105×9[112−142] cm−1 =9.873×105×1516 =925593.75 cm−1 ⇒λ=1.0803×10−6 cm or 1080 nm

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