(solved)Question 2.33 of NCERT Class XI Chemistry Chapter 2

He⁺ is hydrogen-like with 1 electron

wavenumbers in the spectra of He⁺ is

$\displaystyle \overline{\nu} = \frac{R_H}{hc}\bigg(\frac{1}{n_i^2} - \frac{1}{n_f^2}\bigg)Z^2$

Given $\displaystyle n_i = 4 \text{, } n_f = 2\text{, Z for He}^+= 2$

$ \begin{aligned} \therefore \overline{\nu}_{He} &= \frac{R_H}{hc}\bigg(\frac{1}{4^2} - \frac{1}{2^2}\bigg)\times 2^2\\\\ &= \frac{R_H}{hc}\bigg(\frac{1}{4^2} \times 2^2 - \frac{1}{2^2} \times 2^2\bigg)\\\\ &= \frac{R_H}{hc}\bigg(\frac{1}{2^2} - \frac{1}{1^2} \bigg)\\\\ &\text{comparing with hydrogen}\\\\ &\equiv\frac{R_H}{hc}\bigg(\frac{1}{n_i^2} - \frac{1}{n_f^2}\bigg)\\\\ &= \overline{\nu}_H \text{ if } n_i = 2\text{, } n_f = 1 \end{aligned} $

$\displaystyle \therefore \overline{\nu}_{He} = \overline{\nu}_{H} \text{ if transition from } 2 \text{ to } 1$

but $\displaystyle \overline{\nu} = \frac{1}{\lambda}$

$\displaystyle \therefore \lambda_{He} = \lambda_{H} \text{ if transition from } 2 \text{ to } 1$