Question 57: Atoms & Nuclei

Question

The wavelength of the second line of Balmer series is 486.4 nm. What is the wavelength of the first line of Lyman series?

Accepted Answer

Wavelength in Balmer series for hydrogen are given by
$$\begin{aligned}
\frac{1}{\lambda} & =R_{\mathrm{H}}\left(\frac{1}{2^2}-\frac{1}{n^2}ight) \
& =R_{\mathrm{H}}\left(\frac{1}{4}-\frac{1}{n^2}ight), n=3,4,5, \ldots
\end{aligned}$$
The second line in Balmer series corresponds to $$n=4$$
Hence, $$\frac{1}{\lambda_2}=R_{\mathrm{H}}\left(\frac{1}{4}-\frac{1}{16}ight)=\frac{3 R_{\mathrm{H}}}{16}$$
or, $$\lambda_2=\frac{16}{3 R_{\mathrm{H}}}$$
The wavelength of the first line $$(n=2)$$ in Lyman series is
$$\begin{array}
& \frac{1}{\lambda_1}=R_{\mathrm{H}}\left(1-\frac{1}{2^2}ight)=R_{\mathrm{H}}\left(1-\frac{1}{4}ight)=\frac{3 R_{\mathrm{H}}}{4} \
	ext { or, } \quad \lambda_1=\frac{4}{3 R_{\mathrm{H}}} \
	herefore \quad \frac{\lambda_1}{\lambda_2}=\frac{4}{3 R_{\mathrm{H}}} 	imes \frac{3 R_{\mathrm{H}}}{16}=\frac{1}{4} \
\Rightarrow \quad \lambda_1=\frac{\lambda_2}{4}=\frac{486.4}{4}=121.6 \mathrm{~nm}
\end{array}$$

Answer

78.8 nm

Answer

121.6 nm

Answer

418.2 nm

Answer

610.5 nm

Atoms & Nuclei

The wavelength of the second line of Balmer series is 486.4 nm. What is the wavelength of the first line of Lyman series?

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