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) $4\lambda$
Concept. The shortest wavelength of a series is its series limit ($n_2 \to \infty$): $\dfrac{1}{\lambda_{\min}} = \dfrac{R}{n_1^2}$, so $\lambda_{\min} \propto n_1^2$.
Step 1. Lower levels. Balmer: $n_1 = 2$. Brackett: $n_1 = 4$.
Step 2. Ratio.
$$\frac{\lambda_{\text{Brackett}}}{\lambda_{\text{Balmer}}} = \frac{4^2}{2^2} = 4 \;\Rightarrow\; \lambda_{\text{Brackett}} = 4\lambda$$
Answer. $4\lambda$ (Option B).
Solution by Sreeraj P, M.Sc Physics