The ratio of the shortest wavelengths of Brac | vedclass

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

Question

(C) The wavelength $\lambda$ for a hydrogen atom transition is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{

Vedclass · Accepted Answer

$4: 1$

Vedclass · Answer

(C) The wavelength $\lambda$ for a hydrogen atom transition is given by the Rydberg formula: $\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.For the shortest wavelength,the transition occurs from $n_2 = \infty$ to $n_1$.Thus,$\frac{1}{\lambda} = \frac{R}{n_1^2}$,or $\lambda = \frac{n_1^2}{R}$.For the Balmer series,$n_1 = 2$,so $\lambda_{Balmer} = \frac{2^2}{R} = \frac{4}{R}$.For the Bracket series,$n_1 = 4$,so $\lambda_{Bracket} = \frac{4^2}{R} = \frac{16}{R}$.The ratio of the shortest wavelength of the Bracket series to the Balmer series is $\frac{\lambda_{Bracket}}{\lambda_{Balmer}} = \frac{16/R}{4/R} = \frac{16}{4} = 4:1$.

The ratio of the shortest wavelengths of Bracket and Balmer series of hydrogen atom is

Similar Questions

The first member of the Paschen series in the hydrogen spectrum has a wavelength of $18,800 \,\mathring{A}$. The short wavelength limit of the Paschen series is ...... $\mathring{A}$.

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

If ${\lambda _{\max }}$ is $6563\,\mathring{A}$ for the Balmer series of a particular atom,then the wavelength of the second line for the Balmer series will be:

In the spectrum of hydrogen,the ratio of the longest wavelength in the Lyman series to the longest wavelength in the Balmer series is

The spectral series of the hydrogen atom that lies in the visible region of the electromagnetic spectrum is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module