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Q.
What is the ratio of the shortest wavelength of the Balmer series to the shortest wavelength of the Lyman series ?
Atoms
Solution:
For a Balmer series
$\frac{1}{\lambda_{B}} = R\left[\frac{1}{2^{2}} -\frac{1}{n^{2}}\right]\quad...\left(i\right) $
where $n = 3, 4 ,.....$
By putting $n = \infty$ in equation $(i)$, we obtain the series limit of the Balmer series. This is the shortest wavelength of the Balmer series.
or $\lambda_B = \frac{4}{R} \quad...(ii)$
For a Lyman series
$\frac{1}{\lambda_L} = R [\frac{1}{1^2} -\frac{1}{n^2}]\quad ...(iii)$
where $n = 2,3,4,....$
By putting $n = \infty$ in equation $(iii)$, we obtain the series limit of the Lyman series. This is the shortest wavelength of the Lyman series.
or $\lambda_L = \frac{1}{R} \quad...(iv)$
Dividing $(ii)$ by $(iv)$, we get
$\frac{\lambda_B}{\lambda_L} = \frac{4}{1}$