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Question

Chemistry Question on Structure of atom

Which transition in the hydrogen spectrum would have the same wavelength as the Balmer type transition from n=4n =4 to n=2n =2 of He+He ^{+}spectrum

A

n=1n =1 to n=3n =3

B

n=1n =1 to n=2n =2

C

n=2n =2 to n=1n =1

D

n=3n =3 to n=4n =4

Answer

n=2n =2 to n=1n =1

Explanation

Solution

He+ion :
λ(H)1​=R(1)2[n12​1​−n22​1​]
λ(He+)1​=R(2)2[221​−421​]
Given λ(H)=λ(He+)
R(1)2[n12​1​−n22​1​]=R(4)[221​−421​]
n12​1​−n22​1​=121​−221​
On comparing n1​=1&n2​=2