PYQ NEET- Atoms And Nuclei L-2

In hydrogen spectrum, the shortest wavelength in the Balmer series is $\lambda$. The shortest wavelength in the Brackett series is :#

A) $4 \lambda$

B) $9 \lambda$

C) $16 \lambda$

D) $2 \lambda$

Answer: (A) $4 \lambda$#

Sol:#

Shortest wavelength in Balmer series when transition of $e^{-}$from $\infty$ to $\mathrm{n}=2$ $$ \begin{aligned} & \because \frac{1}{\lambda}=\mathrm{Rz}^2\left[\frac{1}{2^2}-\frac{1}{\infty^2}\right] \ & \frac{1}{\lambda}=\frac{R}{4} \ldots(1) \end{aligned} $$

Shortest wavelength is Lyman series when transition of $e^{-}$from $\infty$ to $\mathrm{n}=1$ $$ \frac{1}{\lambda^{\prime}}=\mathrm{R}(1)^2\left[\frac{1}{4^2}-\frac{1}{\infty^2}\right] \Rightarrow \frac{1}{\lambda^{\prime}}=\frac{\mathrm{R}}{16} \ldots . . $$

Eq. (1) / Eq. (2) $$ \frac{\lambda^{\prime}}{\lambda}=\frac{\mathrm{R}}{4} \times \frac{16}{\mathrm{R}} \Rightarrow \lambda^{\prime}=4 \lambda $$