Question 12: Atoms & Nuclei

Question

Which of the following transitions of $ \mathrm{He}^{+} $ion will give rise to spectral line which has same wavelength as the spectral line in hydrogen atom?

Accepted Answer

To determine which transitions of a $\mathrm{He}^{+}$ ion can produce a spectral line with the same wavelength as a given transition in a hydrogen atom, let's examine the Rydberg formula:
For a hydrogen atom ($\mathrm{Z}=1$), the wavelength $\lambda_1$ for a transition from $n_2$ to $n_1$ is given by:
$ \frac{1}{\lambda_1} = R \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) $
Here, $R$ is the Rydberg constant, and $n_1$ and $n_2$ are the principal quantum numbers with $n_2 > n_1$.
For a $\mathrm{He}^{+}$ ion (with $Z=2$), if the electron transitions from $n_4$ to $n_3$, the wavelength $\lambda_2$ is given by:
$ \frac{1}{\lambda_2} = R \cdot 4 \left(\frac{1}{n_3^2} - \frac{1}{n_4^2}\right) $
Assuming $\lambda_1 = \lambda_2$, we have:
$ R \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) = R \cdot 4 \left(\frac{1}{n_3^2} - \frac{1}{n_4^2}\right) $
By comparing both sides, we get the relationships:
$ n_1 = \frac{n_3}{2}, \quad n_2 = \frac{n_4}{2} $
Both $n_1$, $n_2$, $n_3$, and $n_4$ must be integers, as they represent principal quantum numbers.
This condition is satisfied when $n_3 = 2$ and $n_4 = 4$, ensuring that transitions corresponding to these quantum numbers yield the same wavelength for both $\mathrm{He}^{+}$ and hydrogen.

Answer

$ n=4 $ to $ n=2 $

Answer

$ n=6 $ to $ n=5 $

Answer

$ n=6 $ to $ n=3 $

Answer

None of these

Atoms & Nuclei

Which of the following transitions of \mathrm{He}^{+} ion will give rise to spectral line which has same wavelength as the spectral line in hydrogen atom?

Choose the correct answer: