One of the lines in the emission spectrum of Li2+ has the same wavelength as that of 2nd line of Balmer series in hydrogen spectrum. The electronic transition corresponding to this line is :-
A
n=12→n=6
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B
n=4→n=2
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C
n=8→n=2
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D
n=8→n=4
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Solution
The correct option is An=12→n=6 From Rydberg's formula,
1λ=RZ2⎛⎝1n2f−1n2i⎞⎠
For 2nd line of Balmer series in hydrogen spectrum,
Z=1,nf=2,ni=4
⇒1λ=R(1)2(122−142)=3R16
Now, for Li2+, considering option (D), Z=3,n1=6,n2=12