Question

The wavelength of certain electronic transition in hydrogen atom is 91.2 nm. Calculate the corresponding wavelength of $H{e}^{+}$ atom:

• A. 22.8 nm
• B. 45.60 nm
• C. 91.20 nm
• D. 32 nm

Answer 1

The correct option is A 22.8 nm

For hydrogen like atom:
$\frac{1}{\lambda }=Z^2{R}_H(\frac{1}{n_1^2}-\frac{1}{n_2^2})$
where Z = atomic number of an element
As given, $n_1$ and $n_2$ for $H$ and $H{e}^{+}$ are same.
For Hydrogen,
$\frac{1}{{\lambda }_H}=(1{)}^2\times R_H(\frac{1}{n_1^2}-\frac{1}{n_2^2}\left)\right(i)$
For $H{e}^{+}$
$\frac{1}{{\lambda }_{H{e}^{+}}}=(2{)}^2\times R_H(\frac{1}{n_1^2}-\frac{1}{n_2^2}\left)\right(ii)$
on dividing eqn (i) by eqn (ii), we get
${\lambda }_{H{e}^{+}}=22.8nm$