The ratio of the longest to shortest waveleng | vedclass

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

Question

(A) For the Brackett series,the wavelength $\lambda$ is given by the Rydberg formula: $\frac{1}{\lambda} = R \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2}

Vedclass · Accepted Answer

$\frac{25}{9}$

Vedclass · Answer

(A) For the Brackett series,the wavelength $\lambda$ is given by the Rydberg formula: $\frac{1}{\lambda} = R \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} ight]$,where $n_1 = 4$ and $n_2 = 5, 6, 7, \dots$ The longest wavelength $(\lambda_{\max})$ corresponds to the transition from $n_2 = 5$ to $n_1 = 4$: $\frac{1}{\lambda_{\max}} = R \left[ \frac{1}{4^2} - \frac{1}{5^2} ight] = R \left[ \frac{1}{16} - \frac{1}{25} ight] = R \left[ \frac{25 - 16}{400} ight] = \frac{9R}{400}$ $\lambda_{\max} = \frac{400}{9R}$ The shortest wavelength $(\lambda_{\min})$ corresponds to the transition from $n_2 = \infty$ to $n_1 = 4$: $\frac{1}{\lambda_{\min}} = R \left[ \frac{1}{4^2} - \frac{1}{\infty^2} ight] = R \left[ \frac{1}{16} - 0 ight] = \frac{R}{16}$ $\lambda_{\min} = \frac{16}{R}$ The ratio of the longest to shortest wavelength is: $\frac{\lambda_{\max}}{\lambda_{\min}} = \frac{400/9R}{16/R} = \frac{400}{9 	imes 16} = \frac{25}{9}$

The ratio of the longest to shortest wavelengths in the Brackett series of hydrogen spectra is

Similar Questions

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

The ratio of the longest to the shortest wavelength emitted in the Paschen series of a hydrogen atom is:

Explain the transition between energy levels in an atom by drawing a line spectra diagram.

If the series limit frequency of the Balmer series is $\nu_{B}$,then the series limit frequency of the Brackett series is

Which spectral series of the hydrogen atom are found in the infrared region?

Vedclass Test Series

Exam Paper Generator

Online Exam Module