Bohr Model Of The Hydrogen Atom MCQ - Practice Questions with Answers

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 nucleaus

of positive charge $Z_e$ (called hydrogen like atom)
Bohr's model is based on the following postulates-
(1). An atom consists of positively charged nucleaus responsible for almost the entrie mass of the assumption is retention
of Rutherford model.
(2). The electrons revive around the nucleaus in certain permitted circular of an infinite radii.
(3). The permitted orbits are those for which the angular momentum of an electron is an intergral muttiple of $\frac{h}{2\pi }$ where $h$ is the Planck's constant If $m$ is the mass and $v$ is the velocity in a permitted orbit of radius $r$, then

$$L=mvr=\frac{nh}{2\mathrm{\Pi }};n=1,2,3\dots \dots \mathrm{\infty }$$

(4). An electron can transition from a non-radiating orbit to another of a lower energy level. In doing so, a photon is emitted whose energy is equal to the energy difference between the two states. Hence, the frequency of the emitted photon is:

$$h_v=E_i-E_f$$

Ei is the energy of the initial state and Ef is the energy of the final state. Also, $\mathrm{Ei}>\mathrm{Ef}$.

Radius of orbit and velocity of electron

Radius of orbit : For an electron around a stationary nucleaus the electrostatics force of attraction provides the neccesary centripetal force. 

ie. $\frac{1}{4\pi {\epsilon }_0}\frac{(Ze)e}{r^2}=\frac{m{v}^2}{r}\cdots$ (i)
also $mvr=\frac{nh}{2\pi }$
From equation (i) and (ii) radius of r orbit

$$\begin{array}{rl}& r_n=\frac{n^2{h}^2}{4{\pi }^{2}kZm{e}^2}=\frac{n^2{h}^2{\epsilon }_0}{mnZ{e}^2}=0.53\frac{n^2}{Z}A(k=\frac{1}{4\pi {\epsilon }_0})\\ & \Rightarrow r_n\propto \frac{n^2}{Z}\\ \Rightarrow & r_n=0.53\frac{n^2}{Z}{A}^0\end{array}$$

Speed of electron:
From the above relations, speed of electron in $n^{th}$ orbit can be calculated as

$$v_n=\frac{2\pi kZ{e}^2}{nh}=\frac{Z{e}^2}{2{\epsilon }_{0}nh}=\left(\frac{c}{137}\right)\frac{Z}{n}=2.2\times {10}^6\frac{Z}{n}m/\sec$$

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