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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 n2n1=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 nm
According to the given data
n1+n2=4....(i)
n2n1=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=4n1=1

We know,
1λ=RZ2[1n211n22]
=1.097×105×9[112142] cm1
=9.873×105×1516
=925593.75 cm1
λ=1.0803×106 cm or 1080 nm

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