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.

List $I$ (Spectral Lines of Hydrogen for transitions from)	List $II$ (Wavelengths $(nm)$)
$A$. $n_2=3$ to $n_1=2$	$I$. $410.2$
$B$. $n_2=4$ to $n_1=2$	$II$. $434.1$
$C$. $n_2=5$ to $n_1=2$	$III$. $656.3$
$D$. $n_2=6$ to $n_1=2$	$IV$. $486.1$

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

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

The ratio of the largest to shortest wavelengths in the Lyman series of hydrogen spectra is

What is the shortest wavelength present in the Paschen series of spectral lines (in $nm$)?

If the series limit frequency of the Lyman series is $v_L$,then the series limit frequency of the Pfund series 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.