Question

The wave number of energy emitted when an electron jumps from $4^{th}$ orbit to $2^{nd}$ orbit in hydrogen is $20.497{\text{cm}}^{-1}$. The wave number of energy for the same transition in ${\text{He}}^{+}$ is-

• A. $81.988{\text{cm}}^{-1}$
• B. $40.994{\text{cm}}^{-1}$
• C. $20.497{\text{cm}}^{-1}$
• D. $5.099{\text{cm}}^{-1}$

Answer 1

The correct option is A $81.988{\text{cm}}^{-1}$

We know the wave number $\left(\overline{\nu }\right)$ is,

$\overline{\nu }=R{Z}^2(\frac{1}{n{{}_1}^2}-\frac{1}{n{{}_2}^2})$

$\Rightarrow \overline{\nu }\propto Z^2$

$\frac{{\overline{\nu }}_2}{{\overline{\nu }}_1}=\frac{Z_2^2}{Z_1^2}$

Here, $Z_1=1;Z_2=2$

${\overline{\nu }}_2=\frac{2^2}{1}\times 20.497$

$=4\times 20.497=81.988{\text{cm}}^{-1}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(A\right)$ is the correct answer.