Question

The wavelength of the photon emitted, when an electron jumps from fourth orbit to second orbit in hydrogen, is $20.397\text{cm}.$ Wavelength of the photon emitted for the same transition in $H{e}^{+}$, is,

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

Answer 1

The correct option is A $5.099{\text{cm}}^{-1}$




$\Rightarrow \frac{1}{\lambda }=R{Z}^2(\frac{1}{n_f^2}-\frac{1}{n_i^2})$
As $n_f$ and $n_i$ are same in both transitions, we can write,
$\Rightarrow \lambda \propto \frac{1}{Z^2}$

Let ${\lambda }_1$ and ${\lambda }_2$ be the wavelengths emitted during the transition of electron from $n=4$ to $n=2$.
Here, ${\lambda }_1=20.36\text{cm}$
$\Rightarrow {\lambda }_1\propto \frac{1}{Z_1^2}.....1$
$\Rightarrow {\lambda }_2\propto \frac{1}{Z_2^2}.....2$

Dividing $1$ by $2$, we get,

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

$\Rightarrow \frac{20.397}{{\lambda }_2}=\frac{2^2}{1^2}\Rightarrow {\lambda }_2=5.095\text{cm}$

Hence, $\left(A\right)$ is the correct answer.