Question

An ionic atom equivalent to hydrogen atom has wavelength equal to $\frac{1}{4}$ of the wavelengths of hydrogen lines. The ion will be

• A. $H{e}^{+}$
• B. $L{i}^{++}$
• C. $N{e}^{9+}$
• D. $N{a}^{10+}$

Answer 1

The correct option is A $H{e}^{+}$

$\text{By Rydberg's formula ,we have -}$

$\begin{array}{c}{R}_H=\text{Rydberg constant}\\ \lambda =\text{wavelength}\\ Z=\text{atomic num.}\end{array}$

$\begin{array}{c}\text{* For a particular member of spectral line}\\ \text{we have -}\\ \begin{array}{cc}\frac{1}{\lambda }& \alpha z^2\Rightarrow \left[\lambda \alpha \frac{1}{z^2}\right]\\ & \Rightarrow \lambda =\frac{k}{z^2}\text{(let)}-\left(1\right)\\ \Rightarrow & {\lambda }_H=k\end{array}\end{array}$

$\begin{array}{c}\text{For any other ionic atom for which}\\ \text{wavelength}=\frac{{\lambda }_H}{4}=\frac{k}{4}-\left(2\right)\\ \text{from eq (1) and (2)}\\ \text{we get}\frac{k}{z^2}=\frac{k}{4}\\ \Rightarrow z^2=4\\ \Rightarrow z=2\left({He}^{+}\right)\text{atom.}\\ \text{Hence, option (A) is correct.}\end{array}$