In hydrogen atom spectra,if the ratio of wave | vedclass

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

Question

(D) According to Bohr's model,the wavelength $\lambda$ of the emitted radiation when an electron jumps from orbit $n_2$ to $n_1$ is given by the Rydbe

Vedclass · Accepted Answer

$0.021$

Vedclass · Answer

(D) According to Bohr's model,the wavelength $\lambda$ of the emitted radiation when an electron jumps from orbit $n_2$ to $n_1$ is given by the Rydberg formula:$\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$For the first line of the Lyman series,$n_1 = 1$ and $n_2 = 2$:$\frac{1}{\lambda_1} = R \left( \frac{1}{1^2} - \frac{1}{2^2} ight) = R \left( 1 - \frac{1}{4} ight) = \frac{3R}{4} \implies \lambda_1 = \frac{4}{3R}$For the first line of the Balmer series,$n_1 = 2$ and $n_2 = 3$:$\frac{1}{\lambda_2} = R \left( \frac{1}{2^2} - \frac{1}{3^2} ight) = R \left( \frac{1}{4} - \frac{1}{9} ight) = R \left( \frac{9-4}{36} ight) = \frac{5R}{36} \implies \lambda_2 = \frac{36}{5R}$The ratio of the wavelengths is given as $\frac{\lambda_1}{\lambda_2} = 9 \alpha$:$\frac{\lambda_1}{\lambda_2} = \left( \frac{4}{3R} ight) 	imes \left( \frac{5R}{36} ight) = \frac{4 	imes 5}{3 	imes 36} = \frac{20}{108} = \frac{5}{27}$Given $\frac{\lambda_1}{\lambda_2} = 9 \alpha$,we have:$9 \alpha = \frac{5}{27} \implies \alpha = \frac{5}{27 	imes 9} = \frac{5}{243} \approx 0.02057 \approx 0.021$

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

Similar Questions

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

In an $H$-like atom,when an electron transits from energy state $n=5$ to $n=2$,a photon of wavelength $434 \, nm$ is emitted. What will be the wavelength of the photon emitted when the transition occurs from energy state $n=4$ to $n=2$?

$A$ hydrogen atom is excited from the ground state to another state with a principal quantum number equal to $4$. The number of spectral lines in the emission spectra will be:

The smallest wavelength of the Lyman series is $91 \ nm$. The difference between the largest wavelengths of the Paschen and Balmer series is nearly . . . . . . $nm$.

In hydrogen spectrum,which of the following spectral series lies in the ultraviolet region?

Vedclass Test Series

Exam Paper Generator

Online Exam Module