The shortest wavelength in the Paschen series | vedclass

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

Question

(C) The wavelength $\lambda$ for the hydrogen spectrum is given by the Rydberg formula: $\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2

Vedclass · Accepted Answer

$820.4$

Vedclass · Answer

(C) The wavelength $\lambda$ for the hydrogen spectrum is given by the Rydberg formula: $\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.For the Paschen series,the transition occurs to the $n_1 = 3$ energy level.The shortest wavelength corresponds to the transition from the highest energy level,$n_2 = \infty$.Substituting these values into the formula:$\frac{1}{\lambda_S} = R_H \left( \frac{1}{3^2} - \frac{1}{\infty^2} ight) = R_H \left( \frac{1}{9} - 0 ight) = \frac{R_H}{9}$.Therefore,$\lambda_S = \frac{9}{R_H}$.Given $R_H = 1.097 	imes 10^7 \ m^{-1}$,we have:$\lambda_S = \frac{9}{1.097 	imes 10^7} \ m \approx 8.204 	imes 10^{-7} \ m$.Converting to nanometers $(1 \ nm = 10^{-9} \ m)$:$\lambda_S = 820.4 \ nm$.

The shortest wavelength in the Paschen series of the Hydrogen spectrum is (Rydberg constant of hydrogen $R_H = 1.097 \times 10^7 \ m^{-1}$). (in $nm$)

Similar Questions

Explain emission line spectra and absorption spectra.

The shortest wavelength in the Balmer series of a hydrogen atom is equal to the shortest wavelength in the Brackett series of a hydrogen-like atom of atomic number $Z$. The value of $Z$ is:

The ratio of maximum to minimum wavelength in the Balmer series of the hydrogen atom is

The first member of the Balmer series of a hydrogen atom has a wavelength of $6561 \; \mathring{A}$. The wavelength of the second member of the Balmer series (in $nm$) is

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module