Question 425: Atoms & Nuclei

Question

Ratio of longest wavelength corresponding to Lyman and Balmer series in hydrogen spectrum is

Accepted Answer

Wavelength for Lyman series is,
$$\frac{1}{\lambda}=R\left[\frac{1}{1^2}-\frac{1}{\mathrm{n}^2}ight]$$
For the longest wavelength, $$\lambda=\lambda_{\max }$$ and $$\mathrm{n}=2$$
$$\frac{1}{\lambda_{\max (\mathrm{L})}}=\mathrm{R}\left[\frac{1}{1^2}-\frac{1}{2^2}ight]=\frac{3}{4}$$
Wavelength for Balmer series is,
$$\frac{1}{\lambda}=R\left[\frac{1}{2^2}-\frac{1}{n^2}ight]$$
For the longest wavelength, $$\mathrm{n}=3$$
$$\begin{aligned}
& \frac{1}{\lambda_{\max (\mathrm{B})}}=\mathrm{R}\left[\frac{1}{2^2}-\frac{1}{3^2}ight]=\frac{5}{36} \
& \frac{\lambda_{\max (\mathrm{L})}}{\lambda_{\max (\mathrm{B})}}=\frac{4}{3} 	imes \frac{5}{36}=\frac{5}{27}
\end{aligned}$$

Answer

$$\frac{7}{29}$$

Answer

$$\frac{9}{31}$$

Answer

$$\frac{5}{27}$$

Answer

$$\frac{3}{23}$$

Atoms & Nuclei

Ratio of longest wavelength corresponding to Lyman and Balmer series in hydrogen spectrum is

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