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Question

The value of (n2+n1) and (n22n21) for He+ ion in atomic spectrum are 4 and 8 respectively. The wavelength of emitted photon when electron jump from n2 to n1 is:

A
329RH
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B
932RH
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C
932RH
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D
329RH
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Solution

The correct option is C 932RH
Given that
n2+n1=4 .....(1)
n22n21=8 -----(2)
Here,
n1 and n2 are the energy levels of He+ ion.

Solving equation (2), we have,

n22n21=(n2+n1)(n2n1)=8 -----(3)

Substituting (1) in equation (3), we have,

4×(n2n1)=8
(n2n1)=84=2 ------- (4)

Solving equations (1) and (4), i.e.
n2+n1=4 .....(1)
n2n1=2 .....(4)

We get,
n1=1,n2=3

Now, using Rydberg's wavelength equation, we have,

1λ=RHZ2(1n211n22)

where,
λ= wavelength of the emitted photon
RH= Rydberg constant
Z= atomic number of H or H-like ion and
n1 & n2 are the energy levels of H or H-like ion

1λ=RH×22×(112132)
1λ=RH×22×(1119)
1λ=RH×22×(919)
1λ=RH×4×89=32RH9
λ=932RH

(C) option is correct.


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