For any given series of spectral lines of ato | vedclass

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

Question

(C) The wavenumber is given by $\bar{
u} = R_H \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} ight]$.For any series,$\Delta \bar{
u} = \bar{
u}_{\max}

Vedclass · Accepted Answer

$9:4$

Vedclass · Answer

(C) The wavenumber is given by $\bar{
u} = R_H \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} ight]$.For any series,$\Delta \bar{
u} = \bar{
u}_{\max} - \bar{
u}_{\min} = R_H \left[ \frac{1}{n_1^2} - \frac{1}{n_1+1^2} ight] - R_H \left[ \frac{1}{n_1^2} - \frac{1}{\infty^2} ight] = R_H \left[ \frac{1}{\infty^2} - \frac{1}{(n_1+1)^2} ight] = \frac{R_H}{(n_1+1)^2}$.For the Lyman series,$n_1 = 1$,so $\Delta \bar{
u}_{	ext{Lyman}} = \frac{R_H}{(1+1)^2} = \frac{R_H}{4}$.For the Balmer series,$n_1 = 2$,so $\Delta \bar{
u}_{	ext{Balmer}} = \frac{R_H}{(2+1)^2} = \frac{R_H}{9}$.Therefore,the ratio is $\frac{\Delta \bar{
u}_{	ext{Lyman}}}{\Delta \bar{
u}_{	ext{Balmer}}} = \frac{R_H/4}{R_H/9} = \frac{9}{4}$.

For any given series of spectral lines of atomic hydrogen,let $\Delta \bar{\nu} = \Delta \bar{\nu}_{\max} - \Delta \bar{\nu}_{\min}$ be the difference in maximum and minimum wavenumbers in $cm^{-1}$. The ratio $\Delta \bar{\nu}_{\text{Lyman}} / \Delta \bar{\nu}_{\text{Balmer}}$ is

Similar Questions

$A$ surface ejects electrons when hit by green light but not when hit by yellow light. Will electrons be ejected if the surface is hit by red light?

In the Bohr model of an atom, the radius of a hydrogen atom in the ground state is $r_1$ and the radius of a $He^{+}$ ion in the ground state is $r_2$. Which of the following is correct?

Write down the postulates of Bohr's model for the hydrogen atom.

The wavelength of the spectral line obtained in the spectrum of $Li^{2+}$ ion, when the transition takes place between two levels whose sum is $4$ and difference is $2$, is

Vedclass Test Series

Exam Paper Generator

Online Exam Module