The extreme wavelengths of the Paschen series | vedclass

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(B) For the Paschen series,the Rydberg formula is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$,where $n_1 = 3$ an

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$0.818\,\mu m$ and $1.89\,\mu m$

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(B) For the Paschen series,the Rydberg formula is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$,where $n_1 = 3$ and $n_2 = 4, 5, 6, \dots, \infty$.For the maximum wavelength (minimum energy transition),we take $n_1 = 3$ and $n_2 = 4$:$\frac{1}{\lambda_{\max}} = R \left( \frac{1}{3^2} - \frac{1}{4^2} ight) = R \left( \frac{1}{9} - \frac{1}{16} ight) = R \left( \frac{16 - 9}{144} ight) = \frac{7R}{144}$.$\lambda_{\max} = \frac{144}{7R} = \frac{144}{7 	imes 1.097 	imes 10^7 \, 	ext{m}^{-1}} \approx 1.875 	imes 10^{-6} \, 	ext{m} \approx 1.89 \,\mu m$.For the minimum wavelength (maximum energy transition),we take $n_1 = 3$ and $n_2 = \infty$:$\frac{1}{\lambda_{\min}} = R \left( \frac{1}{3^2} - \frac{1}{\infty^2} ight) = \frac{R}{9}$.$\lambda_{\min} = \frac{9}{R} = \frac{9}{1.097 	imes 10^7 \, 	ext{m}^{-1}} \approx 0.820 	imes 10^{-6} \, 	ext{m} \approx 0.818 \,\mu m$.Thus,the extreme wavelengths are $0.818 \,\mu m$ and $1.89 \,\mu m$.

The extreme wavelengths of the Paschen series are

Similar Questions

Assertion : In Lyman series,the ratio of minimum and maximum wavelength is $\frac{3}{4}$.
Reason : Lyman series constitute spectral lines corresponding to transition from higher energy to ground state of hydrogen atom.

Ratio of longest wavelengths corresponding to Lyman and Balmer series in hydrogen spectrum is :-

If $\Delta \lambda_L$ is the difference between the shortest and longest wavelengths of the Lyman series and $\Delta \lambda_B$ is the difference between the shortest and longest wavelengths of the Balmer series,then $\frac{\Delta \lambda_B}{\Delta \lambda_L} = $

The ratio of the longest wavelengths corresponding to the Lyman and Balmer series in the hydrogen spectrum is:

Which of the following statements is true?

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