Question
  1. State Bohr’s postulate to define stable orbits in hydrogen atom. How does de Broglie’s hypothesis explain the stability of these orbits?
  2. A hydrogen atom initially in the ground state absorbs a photon which excites it to the n = 4 level. Estimate the frequency of the photon.

Answer

  1. Quantum condition:
The electrons are permitted to circulate only in those orbits in which the angular momentum of an electron is an integralmultiple of $\frac{\text{h}}{2\pi};$ h being Planck’s constant. Therefore, for any permitted orbit.

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

where L,m and v are the angular momentum,mass and speed of the electron, 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.

Stationary orbits:

While revolving in the permissible orbits, an electron does not radiate energy. These non-radiating orbits are called stationary orbits.

According to de-Broglie, the circumference of $n^{th}$ circular orbit in which electron is revolving is given by:

$2\pi\text{r}_{\text{n}}=\text{n}\lambda\ .....(\text{i})$

where $\lambda=$ wavelength of wave emitted by electron

also, by de-Broglie’s hypothesis

$\lambda=\frac{\text{h}}{\text{mv}}\ ....(\text{ii})$

From (i) and (ii)

$2\pi\text{r}_{\text{n}}=\frac{\text{nh}}{\text{mv}}$

$\therefore\ \text{mvr}_{\text{n}}=\frac{\text{nh}}{2\pi}$
  1. $\frac{1}{\lambda}=\text{R}\Big(\frac{1}{\text{n}^2_1}-\frac{1}{\text{n}^2_2}\Big)$
Here:

$n_1 = 1$

$n_2 = 4$

$\frac{1}{\lambda}=\text{R}\Big(1-\frac{1}{4}\Big)$

$\frac{1}{\lambda}=\frac{3}{4}\text{R}$

$\lambda=\frac{4}{3\text{R}}$

Now, $\text{c}=\text{f}\lambda$

$\text{f}=\frac{\text{c}}{\lambda}=\frac{4\text{c}}{3\text{R}}$

Where c = speed of light in vacuum $= 3 \times 10^8m/ s.$

R = Rhydberg constant $= 1.09 \times 10^7m^{-1}.$

$\text{f}=\frac{4\times3\times10^8}{3\times1.09\times10^7}$

$\text{f}=36.69\text{Hz}$

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