Q 12-12-007NEETNEET 2023Top questionMedium
In hydrogen spectrum, the shortest wavelength in the Balmer series is $\lambda$. The shortest wavelength in the Bracket series is :
Answer: (B) $4\lambda$
The shortest wavelength of a series is the series limit ($n_2 \to \infty$): $\dfrac{1}{\lambda} = \dfrac{R}{n_1^2}$, so $\lambda_{min} \propto n_1^2$.
Balmer: $n_1 = 2$. Brackett: $n_1 = 4$.
$$\frac{\lambda_{\text{Brackett}}}{\lambda_{\text{Balmer}}} = \frac{16}{4} = 4 \;\Rightarrow\; \lambda_{\text{Brackett}} = 4\lambda$$
Solution by Sreeraj P, M.Sc Physics