Question 615: Atoms & Nuclei

Question

The ratio of minimum wavelengths of Balmer and Paschen series of hydrogen atom will be

Accepted Answer

Minimum wavelength for the Balmer series:
The formula for the minimum wavelength $\lambda_{\min B}$ in the Balmer series is given by:
$ \frac{1}{\lambda_{\min B}} = R \left(\frac{1}{2^2} - \frac{1}{\infty^2}\right) $
Simplifying this, we have:
$ \frac{1}{\lambda_{\min B}} = \frac{R}{4} $
Thus, the minimum wavelength for the Balmer series $\lambda_{\min B}$ is:
$ \lambda_{\min B} = \frac{4}{R} $
Minimum wavelength for the Paschen series:
Similarly, for the Paschen series, the formula for the minimum wavelength $\lambda_{\min P}$ is:
$ \frac{1}{\lambda_{\min P}} = R \left(\frac{1}{3^2} - \frac{1}{\infty^2}\right) $
This simplifies to:
$ \frac{1}{\lambda_{\min P}} = \frac{R}{9} $
So, the minimum wavelength for the Paschen series $\lambda_{\min P}$ is:
$ \lambda_{\min P} = \frac{9}{R} $
Calculating the ratio:
Now, to find the ratio of the two wavelengths $\lambda_{\min B}$ and $\lambda_{\min P}$, we use the formulas derived:
$ \frac{\lambda_{\min B}}{\lambda_{\min P}} = \frac{\frac{4}{R}}{\frac{9}{R}} = \frac{4}{9} $
Therefore, the ratio of the minimum wavelengths of the Balmer to the Paschen series is $4:9$.

Answer

$1: 4$

Answer

$4: 9$

Answer

$9: 4$

Answer

$4: 1$

Atoms & Nuclei

The ratio of minimum wavelengths of Balmer and Paschen series of hydrogen atom will be

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