Question

Wavelength of high energy transition of $H$-atom is $91.2nm$. Calculate the corresponding wavelength of $H{e}^{+}$ ion.

• A. $91.2nm$
• B. $22.8nm$
• C. $45.6nm$
• D. $182.4nm$
• E. $364.8nm$

Answer 1

The correct option is B $22.8nm$

For $H$-atom:$\frac{1}{{\lambda }_H}=R_H\cdot [\frac{1}{n_1^2}-\frac{1}{n_2^2}]$
For $H{e}^{+}$ ion:$\frac{1}{{\lambda }_{H{e}^{+}}}=R_H\times Z^2[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$
${\lambda }_{H{e}^{+}}={\lambda }_H\times \frac{1}{Z^2}$
$\because {\lambda }_H=91.2nm,Z$ for $H{e}^{+}=2$
$\therefore {\lambda }_{H{e}^{+}}=91.2\times \frac{1}{2^2}=22.8nm$