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Question

Calculate the energy of radiation emitted for the electronic transition from infinity to ground state for Li2+ ion.
(Given:c=3×108 m/s, RH=1.09678×107 m1
h=6.626×1034 J s1)

A
19.66×1018 J
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B
17.66×1018 J
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C
9.83×1018 J
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D
8.83×1018 J
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Solution

The correct option is A 19.66×1018 J
From Rydberg equation, wavenumber
v=1λ=RHZ2[1n211n22]

when n1=1, n2=, Z=3
1λ=1.09678×107×32[11212]

λ=1.01×108 m=101o A

Energy associated with this radiation,
E=hcλ=6.626×1034×3×108101×1010

=19.66×1018 J

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