Question

If the shortest wave - length of a spectral line in Lyman series for $H{e}^{+}$ is 'x', then what will be the largest wave length of a spectral line in Balmer series of $L{i}^{2+}$

• A. $\frac{4x}{5}$
• B. $9x$
• C. $\frac{5x}{4}$
• D. $\frac{16x}{5}$

Answer 1

The correct option is D

$\frac{16x}{5}$

For the shortest wave length of spectral line in $H{e}^{+}$ the transition should be from $n_{\infty }\to n_1$
Thus $\frac{1}{{\lambda }_{min}}=R_H.\frac{Z^2}{n_1^2}$
or $\frac{1}{x}=R_H.\frac{4}{1}[ForH{e}^{+};Z=2,{\lambda }_{min}]$
or $R_H=\frac{1}{4x}$ ...(i)
For the largest wave length in Balmer series of $L{i}^{2+}$ the transition occurs from $n_2=3$ to $n_2=2$ and in such a case we have
$\frac{1}{{\lambda }_{max}}=R_H.Z^2[\frac{1}{(n^2{)}_1}-\frac{1}{n_2^2}]$
or $\frac{1}{{\lambda }_{max}}=R_H.\left(3{)}^2\right[\frac{1}{(2{)}^2}-\frac{1}{(3{)}^2}]$
or $\frac{1}{{\lambda }_{max}}=\frac{1}{4x}\times [\frac{1}{4}-\frac{1}{9}]$ $[R_H=\frac{1}{4x}]$
or $\frac{1}{{\lambda }_{max}}=\frac{9}{4x}\left(\frac{9-4}{36}\right)$
or $\frac{1}{{\lambda }_{max}}=\frac{9}{4x}\times \frac{5}{36}$
or $\frac{1}{{\lambda }_{max}}=\frac{5}{16x}$
or $\frac{1}{{\lambda }_{max}}=\frac{16x}{5}$