Question

The shortest wavelength of $H$ atom in the Lyman series is ${\lambda }_1.$ The longest wavelength in the Balmer series of $H{e}^{+}$ is :-

• A. $\frac{5{\lambda }_1}{9}$
• B. $\frac{27{\lambda }_1}{5}$
• C. $\frac{9{\lambda }_1}{5}$
• D. $\frac{36{\lambda }_1}{5}$

Answer 1

Answer: (C)

As we know $\Delta E=\frac{hc}{\lambda }$

So $\lambda =\frac{hc}{\Delta E}$ for $\lambda$ minimum i.e.

shortest; $\Delta E=$ maximum

for Lyman series $n=1$ for $\Delta E_{\max }$

Transition must be form $n=\infty$ to $n=1$

So $\frac{1}{\lambda }=R_H{Z}^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

$\frac{1}{\lambda }=R_H{Z}^2(1-0)$

$\frac{1}{\lambda }=R\times (1{)}^2\Rightarrow {\lambda }_1=\frac{1}{R}$

For longest wavelength $\Delta E=$ minimum for Balmer series $n=3$ to $n=2$ will have $\Delta E$ minimum

for $H{e}^{+}Z=2$

So $\frac{1}{{\lambda }_2}=R_H\times Z^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

$\frac{1}{{\lambda }_2}=R_H\times 4\left(\frac{1}{4}-\frac{1}{9}\right)$

$\frac{1}{{\lambda }_2}=R_H\times \frac{5}{9}$

${\lambda }_2={\lambda }_1\times \frac{9}{5}$