The wavelength of the first line of Lyman series for hydrogen is identical to that of the second line of Balmer series for some hydrogen like ion x. Calculate energies of the first four levels of x.
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Answer:
Wavelength of first line of Lyman series :
λ
1
1
=RZ
1
2
(1/1
2
−1/2
2
)=3R/4 (for hydrogen Z=1)
Wavelength of second line of Balmer series of element X :
λ
2
1
=RZ
2
2
(1/2
2
−1/4
2
)=(3R/16)Z
2
2
Here, λ
1
=λ
2
⇒Z
2
=2
Thus, energy levels of X (He
+
) are E
n
=−
n
2
13.6Z
2
eV
So, E
1
=−13.6(4)/1
2
=−54.4eV,E
2
=−13.6(4)/2
2
=−13.6eV,E
3
=−13.6(4)/3
2
=−6.04eV,E
4
=−13.6(4)/16=−3.52eV
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