What is the shortest wavelength present in th | vedclass

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

Question

(D) For the Paschen series,the Rydberg formula is given by $\frac{1}{\lambda} = R \left[ \frac{1}{3^2} - \frac{1}{n^2} \right]$,where $n = 4, 5, 6, \d

Vedclass · Accepted Answer

$820$

Vedclass · Answer

(D) For the Paschen series,the Rydberg formula is given by $\frac{1}{\lambda} = R \left[ \frac{1}{3^2} - \frac{1}{n^2} ight]$,where $n = 4, 5, 6, \dots$ To find the shortest wavelength,we consider the limit where $n = \infty$. Substituting $n = \infty$ into the formula: $\frac{1}{\lambda_{\min}} = R \left[ \frac{1}{9} - 0 ight] = \frac{R}{9}$. Thus,$\lambda_{\min} = \frac{9}{R}$. Given the Rydberg constant $R \approx 1.097 	imes 10^7 \ m^{-1}$: $\lambda_{\min} = \frac{9}{1.097 	imes 10^7} \approx 8.204 	imes 10^{-7} \ m$. Converting to nanometers: $\lambda_{\min} \approx 820.4 \ nm \approx 820 \ nm$.

What is the shortest wavelength present in the Paschen series of spectral lines (in $nm$)?

Similar Questions

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

If radiation of all wavelengths from ultraviolet to infrared is passed through hydrogen gas at room temperature,absorption lines will be observed in the:

$A$ hydrogen atom is excited from the ground state to a state with principal quantum number $n = 4$. The number of spectral lines emitted in the emission spectrum is:

What is emission line spectrum?

Vedclass Test Series

Exam Paper Generator

Online Exam Module