Question

The kinetic energy of the electron in an orbit of radius $r$ in hydrogen atom is ($e = $ electronic charge)
A. $\dfrac{{{e^2}}}{r}$
B. $\dfrac{{{e^2}}}{{2r}}$
C. $\dfrac{{{e^2}}}{{4r}}$
D. $\dfrac{{{e^2}}}{{2{r^2}}}$

Answer 1

Hint Following the Bohr’s atomic model we can calculate the velocity of an electron in a hydrogen atom. Then substitute the value of this velocity in the kinetic energy formula.
Formulas used
$\dfrac{{m{v^2}}}{r} = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{e^2}}}{{{r^2}}}$ where $m$ is the mass of the electron, $v$ is its velocity, $r$ is the radius of its orbit and $e$ is its charge.
$T = \dfrac{1}{2}m{v^2}$ where $T$ is the kinetic energy.

Complete step by step solution
 To calculate the K.E of electrons in an orbit of a hydrogen atom, we have to gain a knowledge of Bohr’s atomic model which put forward the following postulates:
The electrons revolve around the nucleus in a circular orbit under the influence of Coulomb force. These discrete orbits are called stationary states.
Corresponding to each of the stationary states, the orbital angular momentum of the electron $mvr$ is equal to an integral multiple of $\hbar $, i.e. $mvr = n\hbar $
Radiation of energy $h\nu $ is either emitted or absorbed when there is a transition of an electron from one energy state to another.
So, $
  mvr = n\hbar \\
   \Rightarrow v = \dfrac{{n\hbar }}{{mr}} \\
$
To maintain the stability of the circular orbit of the electrons, the Coulomb’s force of attraction is balanced by the centrifugal force
$\dfrac{{m{v^2}}}{r} = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{e^2}}}{{{r^2}}}$
$ \Rightarrow {v^2} = \dfrac{{{e^2}}}{{4\pi {\varepsilon _0}mr}}$
The formula for kinetic energy is $\dfrac{1}{2}m{v^2}$
Substituting the value of ${v^2}$ in this expression we get,
$
  T = \dfrac{1}{2}m \times \dfrac{{{e^2}}}{{4\pi {\varepsilon _0}mr}} \\
   \Rightarrow T = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{e^2}}}{{2r}} \\
$
Since the factor $\dfrac{1}{{4\pi {\varepsilon _0}}}$ is a dimensionless quantity, so we can say that the kinetic energy is proportional to $\dfrac{{{e^2}}}{{2r}}$

Hence, the correct option is B.

Note Even though Bohr revolutionized the whole quantum theory with his atomic model, there were some drawbacks to this. The atomic model was primarily for hydrogen atoms and couldn’t elaborate the spectra of multi-electron systems and was unable to predict the intensities of several lines. Wave nature of electrons was also not justified by this model.