BITSAT 2024 Shift 1 Paper

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 ( $Z=1$ ), 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.