Bohr Model Of The Hydrogen Atom - Practice Questions & MCQ

Bohr's Model of hydrogen atom:  

Bohr proposed a model for hydrogen atom which is also applicable for some lighter atoms in which a single electron revolves around a stationary nucleus of positive charge $Z_e$ (called hydrogen-like atom)
Bohr's model is based on the following postulates-
(1). Bohr's first postulate was that an electron in an atom could revolve in certain stable orbits without the emission of radiant energy, contrary to the predictions of electromagnetic theory. According to this postulate, each atom has certain definite stable states in which it can exist, and each possible state has definite total energy. These are called the stationary states of the atom

For electrons revolving in a stable orbit, the necessary centripetal force is provided by the coulomb's force

$$
\frac{m v_n^2}{r_n}=\frac{k z e^2}{r_n^2}
$$

(2) Bohr's second postulate defines these stable orbits. This postulate states that the electron revolves around the nucleus only in those orbits for which the

 angular momentum is some integral multiple of $\frac{h}{2 \pi}$ where h is

the Planck's constant (= $6.6 \times 10^{-34} \mathrm{~J} \mathrm{~s}$ ). Thus the angular momentum (L) of the orbiting electron is quantised. That is

$$
L=m v_n r_n=\frac{n h}{2 \pi} ; n=1,2,3 \ldots \ldots \infty
$$

(3) Bohr's third postulate incorporated into atomic theory the early quantum concepts that had been developed by Planck and Einstein. It states that an electron might make a transition from one of its specified non-radiating orbits to another of lower energy. When it does so, a photon is emitted having energy equal to the energy difference between the initial and final states. The frequency of the emitted photon is then given by

$$
h \nu=E_i-E_f
$$

$E_i$ is the energy of the initial state and $E_f$ is the energy of the final state. Also, $E_i$ > $\mathrm{E}_{\mathrm{f}}$.
$r_n$-radius of the $n$th orbit
$v_n$ - speed of an electron in the nth orbit

Radius of orbit and velocity of the electron

Radius of the orbit: For an electron around a stationary nucleus the electrostatic force of attraction provides the necessary centripetal force. 

ie. $\frac{1}{4 \pi \varepsilon_0} \frac{(Z e) e}{r^2}=\frac{m v^2}{r} \quad \cdots$ (i)
also $m v r=\frac{n h}{2 \pi}$
From equations (i) and (ii) radius of r orbit

$
\begin{aligned}
& r_n=\frac{n^2 h^2}{4 \pi^2 k Z m e^2}=\frac{n^2 h^2 \varepsilon_0}{\pi m Z e^2}=0.53 \frac{n^2}{Z} A \quad\left(k=\frac{1}{4 \pi \varepsilon_0}\right) \\
& \Rightarrow r_n \propto \frac{n^2}{Z} \\
\Rightarrow & r_n=0.53 \frac{n^2}{Z} \text{Å}
\end{aligned}
$
Speed of electron: 

From the above relations, the speed of electrons in $n^{t h}$ orbit can be calculated as

$
v_n=\frac{2 \pi k Z e^2}{n h}=\frac{Z e^2}{2 \varepsilon_0 n h}=\left(\frac{c}{137}\right) \frac{Z}{n}=2.2 \times 10^6 \frac{Z}{n} \mathrm{~m} / \mathrm{sec}
$

where $\left(c=\right.$ speed of light $\left.=3 \times 10^8 \mathrm{~m} / \mathrm{s}\right)$