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Write first and second postulates of Bohr's Atomic Model. Obtain the expression for radius and velocity of stable orbit of electron.###Explain Bohr's two postulates for hydrogen atom.

Vidyadip · Accepted Answer

Bohr's postulates : Bohr proposed three postulates which are following :
(1) Bohr's first postulates : Any electron can revolve in the definite stable states of atom without radiating the emission energy. It is opposite to the electromagnetic principle. According to this postulate, each atom has same definite stable states in which it existed and existed total energy is definite in all possible states. All these possible states are called stable states of atom.
(2) Bohr's second postulate : Second postulate defines to these stable orbit. According to this postulate, electron revolves around the nucleus only in those orbits for which angular momentum is multiple integer of $\frac{h}{2 \pi}$. Where, $h$ Planck's constant $\left(=6.6 \times 10^{-34} Js \right)$. Therefore, L is quantised angular momentum of revolving electron. That is,
$L=\frac{n h}{2 \pi}$ $\ldots (1)$
(3) Bohr's third postulate : According to this postulate, transition of electron can be done from specially given unradiated orbits to second low energy orbits. When it does so a photon is emitted, whose energy is equal to the difference of energy of initial and final states. Frequency of emitted photon is given by following expression :
$h v=E_i-E_f$ $\ldots (2)$
Where, $E _i$ and $E _f$ are the energies of initial and final states, $E _i> E _f$
Bohr's Radius or radius of stable orbits : Let $r_n$ is the radius of $n^{\text {th }}$ orbit of any atom and in which an electron is moving with velocity $v_n$. If mass of electron is $m$ then from the Bohr's second postulate,
$m v_n r_n=n \cdot \frac{h}{2 \pi}$ $\ldots (1)$

From Bohr's first postulate,
$F _e = F _c$
$\therefore \frac{1}{4 \pi \epsilon_0} \cdot \frac{Z e^2}{r_n^2}=\frac{m v^2 n}{r_n}$ $\ldots (2)$
From eqn. (1) $v_n=\frac{n h}{2 \pi m r_n}$
Put value of $v_n$ in eqn. (2)
$\frac{1}{4 \pi \epsilon_0} \cdot \frac{ Z e^2}{r_n^2}=\frac{m}{r_n} \times \frac{n^2 h^2}{4 \pi^2 m^2 r_n^2}$
or $\quad \frac{Z e^2}{\epsilon_o}=\frac{n^2 h^2}{r_n \cdot \pi m}$ $\quad$ $\therefore r_n=\frac{\epsilon_0 n^2 h^2}{\pi m Z e^2}$
or $\quad \underline {r_n=\frac{\epsilon_0 h^2}{\pi m Z e^2} \cdot n^2}$ $\ldots (3)$
Speed of electron in stable orbit
$\because \quad v_n=\frac{n h}{2 \pi m r_n}$
Put the value of $r_n$ from eqn. (3)
$v_n=\frac{n h}{2 \pi m\left[\frac{\epsilon_0 h^2 n^2}{\pi m e^2 Z}\right]} \Rightarrow \underline {v_n=\frac{Z e^2}{2 \epsilon_0 h n}}$

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