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Atoms — Physics STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 SciencePhysicsAtoms3 Marks
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 Plancks 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 Bohrs 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 $-$ Broglies 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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