Question 478: Atoms & Nuclei

Question

In Balmer series, wavelength of the $$2^{	ext {nd }}$$ line is '$$\lambda_1$$' and for Paschen series, wavelength of the $$1^{	ext {st }}$$ line is '$$\lambda_2$$', then the ratio '$$\lambda_1$$' to '$$\lambda_2$$' is

Accepted Answer

For spectral series, $$\frac{1}{\lambda}=R Z^2\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}ight)$$
For the Balmer series, $$\mathrm{n}_1=2$$
The wavelength for $$2^{	ext {nd }}$$ line of the Balmer series is
$$\begin{aligned}
& \frac{1}{\lambda_1}=\mathrm{RZ}^2\left(\frac{1}{2^2}-\frac{1}{4^2}ight) \
& \frac{1}{\lambda_1}=\mathrm{RZ}^2\left(\frac{1}{4}-\frac{1}{16}ight) \
& \frac{1}{\lambda_1}=\mathrm{RZ}^2\left(\frac{3}{16}ight) \Rightarrow \lambda_1=\left[\frac{16}{3}ight]
\end{aligned}$$
For the Paschen series, $$\mathrm{n}_1=3$$
The wavelength for $$1^{	ext {st }}$$ line of the Paschen series is
$$\begin{aligned}
\frac{1}{\lambda_2} & =R Z^2\left(\frac{1}{3^2}-\frac{1}{4^2}ight) \
\frac{1}{\lambda_2} & =R Z^2\left(\frac{1}{9}-\frac{1}{16}ight) \
\lambda_2 & =\frac{144}{7} \
\frac{1}{\lambda_2} & =R Z^2\left(\frac{7}{144}ight) \
	herefore \quad \frac{\lambda_1}{\lambda_2} & =\frac{16}{3} 	imes \frac{7}{144}=\frac{7}{27}
\end{aligned}$$

Answer

$$5: 128$$

Answer

$$5: 81$$

Answer

$$7: 27$$

Answer

$$9: 132$$

Atoms & Nuclei

In Balmer series, wavelength of the $2^{\text {nd }} line is '\lambda_1' and for Paschen series, wavelength of the 1^{\text {st }} line is '\lambda_2', then the ratio '\lambda_1' to '\lambda_2$' is

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