What transition in the hydrogen spectrum would have the same wavelength as the
Balmer transition n = 4 to n= 2 of He^+ spectrum ?
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Answer:
We have to compare wavelength of transition in the H-spectrum with the Balmer transition n=4 to n=2 of He
+
spectrum.
∵λ
H
=λ
He
+
∴R
H
Z
H
2
[
n
1
2
1
−
n
2
2
1
]=R
H
Z
He
+
2
[
2
2
1
−
4
2
1
]
1×[
n
1
2
1
−
n
2
2
1
]=4×(
4
1
−
16
1
)
[
n
1
2
1
−
n
2
2
1
]=4×
16
4−1
[
n
1
2
1
−
n
2
2
1
]=
4
3
If n
1
=1, then n
2
=2,3,...
For first line n
2
=2, n
1
=1
[
1
2
1
−
2
2
1
]=
1
1
−
4
1
=
4
3
Hence, transition n
2
=2 to n
1
=1 will give spectrum of the same wavelength as that of Balmer transition, n
2
=4 to n
1
=2 in He
+
.
Explanation:
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