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Q.
If the shortest wavelength of the spectral line of $H$-atom in the Lyman series is $x$, then the longest wavelength of the line in Balmer series of $Li ^{2+}$ is
Structure of Atom
Solution:
$\frac{1}{\lambda}= R _{ H } Z ^2\left(\frac{1}{n_1{ }^2}-\frac{1}{n_2^2}\right)$
To calculate shortest wavelength take $n_2=\infty$ and longest wavelength take nearest value of $n_2$.
For $H$-atom, to calculate
$\frac{1}{\lambda_{\text {shortest }}}, n_2=\infty, Z=1, n_1=1$
$ \therefore \frac{1}{x}= R _{ H }$ (Lyman series)
For $\frac{1}{\lambda_{\text {longest }}}$ for $Li ^{2+}, Z=3, n_1=2, n_2=3$ (Balmer series)
$\frac{1}{\lambda_{\text {longest }}}=\frac{1}{x} \times 3^2\left(\frac{1}{2^2}-\frac{1}{3^2}\right)=\frac{5}{4 x} $
$\therefore \lambda_{\text {longest }}=\frac{4 x}{5}$