Question

(i) State bohr's quantization condition for defining stationary orbits. How does de broglie hypothesis explain the stationary orbits?
(ii) Find the relation between the three wavelengths ${\lambda }_1$,${\lambda }_2$ and ${\lambda }_3$ from the energy level diagram shown below.

Answer 1

(i) Bohr's Quantization Rule:

Of all possible circular orbits allowed by the classical theory, the electrons are permitted to circulate only in those orbits in which the angular momentum of an electron is an integral multiple of $\frac{h}{2\pi }$, where $h$ is Plank's constant.

Therefore, for any permitted orbit,

$L=mvr=\frac{nh}{2\pi }$ ; $n=1,2,3,........$

Where $L$, $m$, and $v$ are the angular momentum, mass and the speed of the electron respectively. $r$ is the radius of the permitted orbit and $n$ is positive integer called principal quantum number.

The above equation is Bohr's famous quantum condition. When an electron of mass $m$ is confined to move on a line of length $l$ with velocity $v$, the de-Broglie wavelength $\lambda$ associated with electron is:

$\lambda =\frac{h}{mv}=\frac{h}{P}$

Where $P$ is Linear momentum

$P=\frac{h}{\lambda }=\frac{h}{2l/n}=\frac{nh}{2l}$

When electron revolves in a circular orbit of radius $r$ then $2l=2\pi r$

Therefore, $P=\frac{nh}{2\pi r}$ or $P$ X $r$ $=\frac{nh}{2\pi }$

(ii) Using Rydberg's formula for spectra of hydrogen atom, we have

$\frac{1}{{\lambda }_1}=R(\frac{1}{n_2^2}-\frac{1}{n_3^2})$ .....(1)

$\frac{1}{{\lambda }_2}=R(\frac{1}{n_1^2}-\frac{1}{n_2^2})$ .....(2)

$\frac{1}{{\lambda }_3}=R(\frac{1}{n_1^2}-\frac{1}{n_3^2})$ .....(3)

Now adding (1) and (2), we get

$\frac{1}{{\lambda }_1}+\frac{1}{{\lambda }_2}=R(\frac{1}{n_1^2}-\frac{1}{n_3^2})=\frac{1}{{\lambda }_3}$

That is, $\frac{1}{{\lambda }_1}+\frac{1}{{\lambda }_2}=\frac{1}{{\lambda }_3}$

Hence, the relation between 3 wavelengths from the energy level diagram is obtained.