The longest wavelength associated with the Pa | vedclass

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Question

(D) The Rydberg formula for the wavelength $\lambda$ of a spectral line is given by: $\frac{1}{\lambda} = R_H \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2}

Vedclass · Accepted Answer

$1.876 	imes 10^{-6} \ m$

Vedclass · Answer

(D) The Rydberg formula for the wavelength $\lambda$ of a spectral line is given by: $\frac{1}{\lambda} = R_H \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} ight]$.For the Paschen series,the transition occurs to the $n_1 = 3$ energy level.The longest wavelength corresponds to the smallest energy difference,which occurs for the transition from the nearest higher energy level,i.e.,$n_2 = 4$.Substituting these values into the formula:$\frac{1}{\lambda} = R_H \left[ \frac{1}{3^2} - \frac{1}{4^2} ight]$$\frac{1}{\lambda} = R_H \left[ \frac{1}{9} - \frac{1}{16} ight] = R_H \left[ \frac{16 - 9}{144} ight] = R_H \left[ \frac{7}{144} ight]$$\lambda = \frac{144}{7 R_H} = \frac{144}{7 	imes 1.097 	imes 10^7}$$\lambda \approx 1.876 	imes 10^{-6} \ m$.

The longest wavelength associated with the Paschen series is: (Given $R_H = 1.097 \times 10^7 \ m^{-1}$)

Similar Questions

In the Balmer series,the wavelength of the $2^{\text{nd}}$ line is $\lambda_1$ and for the Paschen series,the wavelength of the $1^{\text{st}}$ line is $\lambda_2$. Then the ratio $\lambda_1 : \lambda_2$ is:

The ratio of the shortest wavelength of the Balmer series to the shortest wavelength of the Lyman series for a hydrogen atom is:

$A$ $12.5\; eV$ electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?

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