Question 479: Atoms & Nuclei

Question

In Lyman series, series limit of wavelength is $$\lambda_1$$. The wavelength of first line of Lyman series is $$\lambda_2$$ and in Balmer series, the series limit of wavelength is $$\lambda_3$$. Then the relation between $$\lambda_1$$, $$\lambda_2$$ and $$\lambda_3$$ is

Accepted Answer

According to Rydberg's formula,
$$\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{\mathrm{n}^2}-\frac{1}{\mathrm{~m}^2}ight)$$
For series limit of Lyman series,
$$\begin{aligned}
& \mathrm{n}=1, \mathrm{~m}=\infty, \lambda=\lambda_1 \
	herefore \quad & \frac{1}{\lambda_1}=\mathrm{R}
\end{aligned}$$
For $$1^{	ext {st }}$$ line of Lyman series,
$$\begin{aligned}
& \mathrm{n}=1, \mathrm{~m}=2, \lambda=\lambda_2 \
	herefore \quad & \frac{1}{\lambda_2}=\frac{3 \mathrm{R}}{4}
\end{aligned}$$
For series limit of Balmer series,
$$\begin{aligned}
& \mathrm{n}=2, \mathrm{~m}=\infty, \lambda=\lambda_3 \
	herefore \quad & \frac{1}{\lambda_3}=\frac{\mathrm{R}}{4}
\end{aligned}$$
Now, $$\frac{1}{\lambda_1}-\frac{1}{\lambda_2}=\mathrm{R}-\frac{3 \mathrm{R}}{4}=\frac{\mathrm{R}}{4}$$
$$	herefore \quad \frac{1}{\lambda_1}-\frac{1}{\lambda_2}=\frac{1}{\lambda_3}$$

Answer

$$\lambda_1=\lambda_2+\lambda_3$$

Answer

$$\lambda_2=\lambda_1+\lambda_3$$

Answer

$$\frac{1}{\lambda_1}=\frac{1}{\lambda_2}-\frac{1}{\lambda_3}$$

Answer

$$\frac{1}{\lambda_1}-\frac{1}{\lambda_2}=\frac{1}{\lambda_3}$$

Atoms & Nuclei

In Lyman series, series limit of wavelength is $\lambda_1. The wavelength of first line of Lyman series is \lambda_2 and in Balmer series, the series limit of wavelength is \lambda_3. Then the relation between \lambda_1, \lambda_2 and \lambda_3$ is

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