Find the maximum wavelength of the Brackett s | vedclass

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

Question

(C) For the Brackett series,the transition occurs from $n_2$ to $n_1 = 4$.For the maximum wavelength (minimum energy),the transition must be from the

Vedclass · Accepted Answer

$40400$

Vedclass · Answer

(C) For the Brackett series,the transition occurs from $n_2$ to $n_1 = 4$.For the maximum wavelength (minimum energy),the transition must be from the nearest energy level,i.e.,$n_2 = 5$.Using the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.Substituting the values: $\frac{1}{\lambda_{\max}} = R \left( \frac{1}{4^2} - \frac{1}{5^2} ight) = R \left( \frac{1}{16} - \frac{1}{25} ight) = R \left( \frac{25 - 16}{400} ight) = R \left( \frac{9}{400} ight)$.Therefore,$\lambda_{\max} = \frac{400}{9R}$.Given $R \approx 1.097 	imes 10^7 \ m^{-1}$,$\lambda_{\max} = \frac{400}{9 	imes 1.097 	imes 10^7} \approx 4.05 	imes 10^{-6} \ m = 40500 \ \mathring A$.Rounding to the nearest provided option,the correct value is $40400 \ \mathring A$.

Find the maximum wavelength of the Brackett series for a hydrogen atom in $\mathring A$.

Similar Questions

What is emission line spectrum?

What is the wavelength of the least energetic photon emitted in the Lyman series of the hydrogen atom spectrum? (in $nm$)

The ratio of longest wavelengths of the spectral lines in the Lyman and Balmer series of hydrogen spectrum is

The ratio of the frequencies of the long wavelength limits of Lyman and Balmer series of hydrogen spectrum is

The ratio of the largest to shortest wavelengths in the Lyman series of hydrogen spectra is

Vedclass Test Series

Exam Paper Generator

Online Exam Module