In the line spectra of the hydrogen atom,the | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The wavelength $\lambda$ is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$,where $R$ is the Rydberg constant. L

Vedclass · Accepted Answer

$10553$

Vedclass · Answer

(A) The wavelength $\lambda$ is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$,where $R$ is the Rydberg constant. Let $C = \frac{1}{R}$. Then $\lambda = C \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)^{-1}$.For the Lyman series $(n_1 = 1)$:Shortest wavelength $\lambda_{L,s} = C \left( \frac{1}{1^2} - \frac{1}{\infty^2} ight)^{-1} = C$.Longest wavelength $\lambda_{L,l} = C \left( \frac{1}{1^2} - \frac{1}{2^2} ight)^{-1} = \frac{4C}{3}$.Difference $\Delta \lambda_L = \frac{4C}{3} - C = \frac{C}{3} = 304\,\mathring{A} \implies C = 912\,\mathring{A}$.For the Paschen series $(n_1 = 3)$:Shortest wavelength $\lambda_{P,s} = C \left( \frac{1}{3^2} - \frac{1}{\infty^2} ight)^{-1} = 9C$.Longest wavelength $\lambda_{P,l} = C \left( \frac{1}{3^2} - \frac{1}{4^2} ight)^{-1} = C \left( \frac{1}{9} - \frac{1}{16} ight)^{-1} = C \left( \frac{7}{144} ight)^{-1} = \frac{144C}{7}$.Difference $\Delta \lambda_P = \frac{144C}{7} - 9C = \frac{144C - 63C}{7} = \frac{81C}{7}$.Substituting $C = 912\,\mathring{A}$:$\Delta \lambda_P = \frac{81 	imes 912}{7} \approx 10553.14\,\mathring{A}$.

In the line spectra of the hydrogen atom,the difference between the largest and the shortest wavelengths of the Lyman series is $304\,\mathring{A}$. The corresponding difference for the Paschen series in $\mathring{A}$ is:

Similar Questions

The ratio of longest wavelength lines in the Balmer and Paschen series of hydrogen spectrum is

$A$ $12.5\,eV$ electron beam is used to bombard gaseous hydrogen at room temperature. The number of spectral lines emitted will be:

In the hydrogen spectrum,the ratio of the wavelengths for Lyman-alpha radiation to Balmer-alpha radiation is

The number of possible spectral lines in a hydrogen atom is:

The spectral series observed for the hydrogen atom found in the visible region is

Vedclass Test Series

Exam Paper Generator

Online Exam Module