Question

The ratio of wave numbers for the highest energy transition of an electron in Lyman series of $H{e}^{+}$ and Balmer series of H-atom is:

• A. $16:1$
• B. $1:1$
• C. $1:16$
• D. $1:4$

Answer 1

The correct option is A $16:1$

$\text{Lyman series of}H{e}^{+}\frac{1}{{\lambda }_1}={\overline{v}}_1=R_H.Z^2[\frac{1}{1^2}-\frac{1}{n_2^2}]$
where $\lambda$ = wavelength of the transition
Z = atomic number and $n_2$ = higher orbit number

$\frac{1}{{\lambda }_1}={\overline{v}}_1=R_H.2^2[\frac{1}{1^2}-\frac{1}{n_2^2}]$....(1)

$\text{Balmer series of H-atom}$
$\frac{1}{{\lambda }_2}={\overline{v}}_2=R_H.1^2[\frac{1}{2^2}-\frac{1}{n_2^2}]$... (2)
For the highest energy transition:
For Lyman series $n_2=\infty$
For Balmer series $n_2=\infty$
Dividing equation (1) by (2), we get
$\therefore \frac{{\overline{v}}_1}{{\overline{v}}_2}=\frac{16}{1}$