Question 542: Atoms & Nuclei

Question

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

Accepted Answer

Shortest wavelength in Balmer series when transition of $$e^{-}$$ from $$\infty$$ to $$\mathrm{n}=2$$
$$\because \frac{1}{\lambda}=\mathrm{Rz}^{2}\left[\frac{1}{2^{2}}-\frac{1}{\infty^{2}}ight]$$
$$\frac{1}{\lambda}=\frac{R}{4}$$ .... (1)
Shortest wavelength is Bracket series when transition of $$e^{-}$$ from $$\infty$$ to $$\mathrm{n}=4$$
$$\frac{1}{\lambda^{\prime}}=\mathrm{R}(1)^{2}\left[\frac{1}{4^{2}}-\frac{1}{\infty^{2}}ight] \Rightarrow \frac{1}{\lambda^{\prime}}=\frac{\mathrm{R}}{16}$$ ..... (2)
Eq. (1) / Eq. (2)
$$\frac{\lambda^{\prime}}{\lambda}=\frac{\mathrm{R}}{4} 	imes \frac{16}{\mathrm{R}} \Rightarrow \lambda^{\prime}=4 \lambda$$

Answer

$$4 \lambda$$

Answer

$$9 \lambda$$

Answer

$$16 \lambda$$

Answer

$$2 \lambda$$

Atoms & Nuclei

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

Choose the correct answer: