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Question

# The value of (n2+n1) and (n22âˆ’n21) 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 932RHGiven that n2+n1=4 .....(1) n22−n21=8 -----(2)Here, n1 and n2 are the energy levels of He+ ion.Solving equation (2), we have, n22−n21=(n2+n1)(n2−n1)=8 -----(3)Substituting (1) in equation (3), we have, 4×(n2−n1)=8 ⇒(n2−n1)=84=2 ------- (4)Solving equations (1) and (4), i.e. n2+n1=4 .....(1) n2−n1=2 .....(4)We get, n1=1,n2=3Now, using Rydberg's wavelength equation, we have, 1λ=RHZ2(1n21−1n22)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×(112−132) ⇒1λ=RH×22×(11−19) ⇒1λ=RH×22×(9−19) ⇒1λ=RH×4×89=32RH9 ⇒λ=932RH∴ (C) option is correct.

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