Question 448: Atoms & Nuclei

Question

The ratio of wavelengths for transition of electrons from $2^{	ext {nd }}$ orbit to $1^{	ext {st }}$ orbit of Helium $\left(\mathrm{He}^{++}ight)$ and Lithium $\left(\mathrm{Li}^{++1}ight)$ is (Atomic number of Helium $=2$, Atomic number of Lithium $=3$ )

Accepted Answer

To find the ratio of the wavelengths for the transition of electrons from the $2^{	ext{nd}}$ orbit to the $1^{	ext{st}}$ orbit for Helium ion $	ext{He}^{++}$ and Lithium ion $\mathrm{Li}^{+}$, we can use the formula 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} ight) $
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 $2^{	ext{nd}}$ orbit to the $1^{	ext{st}}$ 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} ight) = RZ^2 \left( \frac{3}{4} ight) $
Therefore, $\lambda = \frac{4}{3RZ^2}$.
For Helium ion $	ext{He}^{++}$ ($Z = 2$):
$ \lambda_{	ext{He}} = \frac{4}{3R \cdot 2^2} = \frac{4}{12R} = \frac{1}{3R} $
For Lithium ion $\mathrm{Li}^{+}$ ($Z = 3$):
$ \lambda_{	ext{Li}} = \frac{4}{3R \cdot 3^2} = \frac{4}{27R} $
To find the ratio of the wavelengths ($\lambda_{	ext{He}} : \lambda_{	ext{Li}}$), we divide the expressions obtained for $\lambda_{	ext{He}}$ and $\lambda_{	ext{Li}}$:
$ \frac{\lambda_{	ext{He}}}{\lambda_{	ext{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 $2^{	ext{nd}}$ orbit to the $1^{	ext{st}}$ orbit of Helium ion ($	ext{He}^{++}$) to Lithium ion ($\mathrm{Li}^{+}$) is
$9 : 4$
.
Therefore, the correct answer is
Option A: $9: 4$
.

Answer

$9: 4$

Answer

$9: 36$

Answer

$4: 9$

Answer

$2: 3$

Atoms & Nuclei

The ratio of wavelengths for transition of electrons from 2^{\text {nd }} orbit to 1^{\text {st }} orbit of Helium \left(\mathrm{He}^{++}\right) and Lithium \left(\mathrm{Li}^{++1}\right) is (Atomic number of Helium =2, Atomic number of Lithium =3 )

Choose the correct answer: