The ratio of minimum wavelengths of Lyman and | vedclass

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

Question

(B) The wavelength $\lambda$ is given by the Rydberg formula: $\frac{1}{\lambda} = R_H Z^2 (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.For the minimum wavele

Vedclass · Accepted Answer

$0.25$

Vedclass · Answer

(B) The wavelength $\lambda$ is given by the Rydberg formula: $\frac{1}{\lambda} = R_H Z^2 (\frac{1}{n_1^2} - \frac{1}{n_2^2})$.For the minimum wavelength,the transition occurs from $n_2 = \infty$ to $n_1$.For the Lyman series,$n_1 = 1$,so $\frac{1}{\lambda_L} = R_H (\frac{1}{1^2} - 0) = R_H$,which means $\lambda_L = \frac{1}{R_H}$.For the Balmer series,$n_1 = 2$,so $\frac{1}{\lambda_B} = R_H (\frac{1}{2^2} - 0) = \frac{R_H}{4}$,which means $\lambda_B = \frac{4}{R_H}$.The ratio of minimum wavelengths $\frac{\lambda_L}{\lambda_B} = \frac{1/R_H}{4/R_H} = \frac{1}{4} = 0.25$.

The ratio of minimum wavelengths of Lyman and Balmer series will be $:-$

Similar Questions

If the first emission in the Lyman series of a hydrogen atom occurs at $121.5 \ nm$,what will be the energy difference between the first and second orbits in $kJ/mol$?

The distance between the $4^{th}$ and $3^{rd}$ Bohr orbits of $He^{+}$ is $:-$

Which of the following represents the wavelength of a spectral line of the Balmer series of the $He^{+}$ ion? ($R=$ Rydberg constant,$n > 2$)

Positronium consists of an electron and a positron (a particle which has the same mass as an electron,but opposite charge) orbiting round their common centre of mass. Calculate the value of the Rydberg constant for this system.

The radius of the first Bohr orbit of a hydrogen atom is $0.53 \times 10^{-8} \ cm$. The velocity of the electron in the first Bohr orbit is

Vedclass Test Series

Exam Paper Generator

Online Exam Module