Assertion: In the Lyman series,the ratio of m | vedclass

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

Question

(A) The Rydberg formula is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right)$.For the Lyman series,the ground s

Vedclass · Accepted Answer

If both Assertion and Reason are correct and the Reason is a correct explanation of the Assertion.

Vedclass · Answer

(A) The Rydberg formula is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right)$.For the Lyman series,the ground state is $n_{1} = 1$.For the minimum wavelength (shortest wavelength),$n_{2} = \infty$:$\frac{1}{\lambda_{min}} = R \left( \frac{1}{1^{2}} - \frac{1}{\infty^{2}} \right) = R \implies \lambda_{min} = \frac{1}{R}$.For the maximum wavelength (longest wavelength),$n_{2} = 2$:$\frac{1}{\lambda_{max}} = R \left( \frac{1}{1^{2}} - \frac{1}{2^{2}} \right) = R \left( 1 - \frac{1}{4} \right) = \frac{3R}{4} \implies \lambda_{max} = \frac{4}{3R}$.The ratio of minimum to maximum wavelength is $\frac{\lambda_{min}}{\lambda_{max}} = \frac{1/R}{4/3R} = \frac{3}{4}$.Both the Assertion and Reason are correct,and the Reason correctly explains the origin of the Lyman series transitions.

Assertion: In the Lyman series,the ratio of minimum and maximum wavelength is $\frac{3}{4}$.
Reason: The Lyman series constitutes spectral lines corresponding to transitions from higher energy levels to the ground state of the hydrogen atom.

Similar Questions

$A$ $12.5\,eV$ electron beam is used to bombard gaseous hydrogen at room temperature. The number of spectral lines emitted will be:

If ${\lambda _{{\text{max}}}}$ is $6563 \text{ Å}$, then the wavelength of the second line of the Balmer series will be:

If $\lambda_{1}$ and $\lambda_{2}$ are the wavelengths of the third member of the Lyman series and the first member of the Paschen series respectively,then the value of $\lambda_{1} : \lambda_{2}$ is

The wave number of the first line in the Lyman series of a hydrogen atom is ($R$ is Rydberg's constant).

The first member of the Balmer series of a hydrogen atom has a wavelength of $6561 \; \mathring{A}$. The wavelength of the second member of the Balmer series (in $nm$) is

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Assertion: In the Lyman series,the ratio of minimum and maximum wavelength is $\frac{3}{4}$.Reason: The Lyman series constitutes spectral lines corresponding to transitions from higher energy levels to the ground state of the hydrogen atom.