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Q.
The shortest wavelength in Lyman series is $ 91.2\,nm $. The longest wavelength of the series is
J & K CETJ & K CET 2006Atoms
Solution:
The wavelength $(\lambda)$ of lines is given by
$\frac{1}{\lambda}=R\left(\frac{1}{1^{2}}-\frac{1}{n^{2}}\right)$
For Lyman series, the shortest wavelength is for
$n=\infty$ and longest is for $n=2$
$\therefore \frac{1}{\lambda_{S}}=R\left(\frac{1}{1^{2}}\right) \ldots$ (i)
$\frac{1}{\lambda_{L}}=R\left(\frac{1}{1}-\frac{1}{2^{2}}\right)=\frac{3}{4} R \ldots$ (ii)
Dividing Eq. (ii) by Eq. (i), we get
$\frac{\lambda_{L}}{\lambda_{S}}=\frac{4}{3}$
Given, $\lambda_{S}=91.2 \,n m$
$\Rightarrow \lambda_{L}=91.2 \times \frac{4}{3}$
$=121.6\, nm$