mula derived from the Rydberg equation for wavelengths of emitted photons during electron transitions in hydrogen-like atoms.

The Rydberg formula for the wavelength $\lambda$ is given by:

$ \frac{1}{\lambda} = RZ^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) $

where,

$R$ is the Rydberg constant for hydrogen,

$Z$ is the atomic number of the nucleus,

$n_1$ is the lower energy level,

$n_2$ is the higher energy level.

For a transition from the $2^{\text{nd}}$ orbit to the $1^{\text{st}}$ orbit ($n_2 = 2$ and $n_1 = 1$), the formula simplifies to:

$ \frac{1}{\lambda} = RZ^2 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = RZ^2 \left( \frac{3}{4} \right) $

Therefore, $\lambda = \frac{4}{3RZ^2}$.

For Helium ion $\text{He}^{++}$ ($Z = 2$):

$ \lambda_{\text{He}} = \frac{4}{3R \cdot 2^2} = \frac{4}{12R} = \frac{1}{3R} $

For Lithium ion $\mathrm{Li}^{+}$ ($Z = 3$):

$ \lambda_{\text{Li}} = \frac{4}{3R \cdot 3^2} = \frac{4}{27R} $

To find the ratio of the wavelengths ($\lambda_{\text{He}} : \lambda_{\text{Li}}$), we divide the expressions obtained for $\lambda_{\text{He}}$ and $\lambda_{\text{Li}}$:

$ \frac{\lambda_{\text{He}}}{\lambda_{\text{Li}}} = \frac{\frac{1}{3R}}{\frac{4}{27R}} = \frac{27}{12} = \frac{9}{4} $

Thus, the ratio of the wavelengths for the transition of electrons from the $2^{\text{nd}}$ orbit to the $1^{\text{st}}$ orbit of Helium ion ($\text{He}^{++}$) to Lithium ion ($\mathrm{Li}^{+}$) is $9 : 4$.

Therefore, the correct answer is Option A: $9: 4$.

Atoms And Nuclei

Continue this paper

Similar Atoms And Nuclei PYQs

More practice from this question