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Q.
In the spectrum of hydrogen atom, the ratio of the longest wavelength in Lyman series to the longest wavelength in the Balmer series is
NTA AbhyasNTA Abhyas 2022
Solution:
When an atom comes down from some higher energy level to the first energy level then emitted lines form of Lyman series.
$\frac{1}{\lambda_{L}}=R \frac{1}{1^{2}}-\frac{1}{n^{2}}$
where $R$ is Rydberg's constant.
When an atom comes from higher energy level to the second level, then Balmer series are obtained.
$\frac{1}{\lambda_{B}}=R \frac{1}{2^{2}}-\frac{1}{n^{2}}$
For maximum wavelength
$n=2, \frac{1}{\lambda_{L}}=R 1-\frac{1}{(2)^{2}}=R 1-\frac{1}{4}=\frac{3 R}{4} \ldots .(i)$
$n=3, \frac{1}{\lambda_{B}}=R \frac{1}{(2)^{2}}-\frac{1}{(3)^{2}}=R \frac{5}{36}\ldots .(ii)$
Dividing Eq. (ii) by Eq. (i), we get
$\frac{\lambda_{L}}{\lambda_{B}}=\frac{5}{27}$