Question 375: Atoms & Nuclei

Question

The ratio of wavelength of spectral lines $$\mathrm{H}_{\alpha}$$ and $$\mathrm{H}_{\beta}$$ in the Balmer series is $$\frac{x}{20}$$. The value of $$x$$ is _________.

Accepted Answer

The Balmer series corresponds to electronic transitions in a hydrogen atom that terminate in the second (n=2) energy level. The spectral lines in the Balmer series are often labeled according to a Greek letter scheme, with H$_\alpha$ corresponding to the n=3 to n=2 transition, H$_\beta$ corresponding to the n=4 to n=2 transition, and so on.
The wavelength of a spectral line in the Balmer series can be calculated using the Rydberg formula:
$
\frac{1}{\lambda} = R_H \left( \frac{1}{2^2} - \frac{1}{n^2} \right)
$
where $R_H$ is the Rydberg constant for hydrogen, $n$ is the principal quantum number corresponding to the initial energy level, and $\lambda$ is the wavelength of the spectral line.
Using this formula, the wavelength of the H$_\alpha$ line is:
$
\frac{1}{\lambda_{\alpha}} = R_H \left( \frac{1}{2^2} - \frac{1}{3^2} \right)
$
And the wavelength of the H$_\beta$ line is:
$
\frac{1}{\lambda_{\beta}} = R_H \left( \frac{1}{2^2} - \frac{1}{4^2} \right)
$
Therefore, the ratio of the wavelengths of the H$_\alpha$ and H$_\beta$ lines is:
$
\frac{\lambda_{\alpha}}{\lambda_{\beta}} = \frac{\left( \frac{1}{2^2} - \frac{1}{4^2} \right)}{\left( \frac{1}{2^2} - \frac{1}{3^2} \right)} = \frac{\frac{3}{16}}{\frac{5}{36}} = \frac{27}{20}
$
So, comparing with the given ratio $\frac{x}{20}$, we find that $x = 27$.

Atoms & Nuclei

The ratio of wavelength of spectral lines $\mathrm{H}_{\alpha} and \mathrm{H}_{\beta} in the Balmer series is \frac{x}{20}. The value of x$ is _________.

Choose the correct answer: