The wavelength of the first spectral line in | vedclass

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

Question

(A) The Rydberg formula for the wavelength $\lambda$ of a spectral line is given by $\frac{1}{\lambda} = R Z^2 \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2

Vedclass · Accepted Answer

$1215 \mathring A$

Vedclass · Answer

(A) The Rydberg formula for the wavelength $\lambda$ of a spectral line is given by $\frac{1}{\lambda} = R Z^2 \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} ight]$.For the first spectral line of the Balmer series in a hydrogen atom $(Z=1, n_1=2, n_2=3)$:$\frac{1}{\lambda_1} = R (1)^2 \left[ \frac{1}{2^2} - \frac{1}{3^2} ight] = R \left[ \frac{1}{4} - \frac{1}{9} ight] = R \left( \frac{5}{36} ight) \implies \lambda_1 = \frac{36}{5R} = 6561 \mathring A$.For the second spectral line of the Balmer series in a singly-ionized helium atom $(Z=2, n_1=2, n_2=4)$:$\frac{1}{\lambda_2} = R (2)^2 \left[ \frac{1}{2^2} - \frac{1}{4^2} ight] = 4R \left[ \frac{1}{4} - \frac{1}{16} ight] = 4R \left( \frac{3}{16} ight) = R \left( \frac{3}{4} ight) \implies \lambda_2 = \frac{4}{3R}$.Dividing $\lambda_2$ by $\lambda_1$:$\frac{\lambda_2}{\lambda_1} = \frac{4/3R}{36/5R} = \frac{4}{3} 	imes \frac{5}{36} = \frac{5}{27}$.Therefore,$\lambda_2 = \frac{5}{27} 	imes 6561 \mathring A = 5 	imes 243 \mathring A = 1215 \mathring A$.

The wavelength of the first spectral line in the Balmer series of a hydrogen atom is $6561 \mathring A$. The wavelength of the second spectral line in the Balmer series of a singly-ionized helium atom is

Similar Questions

In the given figure,the energy levels of a hydrogen atom have been shown along with some transitions marked $A, B, C, D$ and $E$. The transitions $A, B$ and $C$ respectively represent:

Which series of the hydrogen atom lie in the infrared region?

Which of the following spectral series of the hydrogen atom lies entirely in the ultraviolet region?

The shortest wavelength of the spectral lines in the Lyman series of the hydrogen spectrum is $915 \text{ Å}$. The longest wavelength of spectral lines in the Balmer series will be: (in $\text{ Å}$)

In hydrogen atom spectra,if the ratio of wavelengths corresponding to the first line of the Lyman series and the first line of the Balmer series is $9 \alpha$,the value of $\alpha$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module