Question

Derive the expression for total energy of electron in ${n^{th}}$ Bohr orbit and show that ${E_n} \propto \dfrac{1}{{{n^2}}}$

Answer 1

Hint: Bohr’s atomic model is a system consisting of electrons revolving around a dense nucleus. The model is based on the Bohr’s postulates. To solve this problem, consider the Bohr’s Postulates. Using these postulates, one can determine the energy of the hydrogen atom for any principal quantum number $n$, the radii of different orbits.

Complete step by step answer:
The Bohr’s Postulates are as follows: The electron revolves around the nucleus at stable orbits without radiating energy. The stable orbits are called the stationary orbits. These orbits are unique and have different radii. The electron cannot orbit in between two stationary orbits, meaning the orbits are uniquely determined.

The stationery orbits are obtained by equating the angular momentum of the electron to integral multiple of reduced Planck’s constant, $mvr = n\dfrac{h}{{2\pi }}$, where v is the velocity of the electron in the ${n^{th}}$ orbit and $r$ is the radius of the ${n^{th}}$ orbit. The orbits have definite energies called the energy shells or energy levels.

Consider a hydrogen like atom having only one electron. The electron is moving with a constant velocity $v$ in an orbit of radius. The force acting on the electron will be given by Coulomb's Law.
$F = \dfrac{{(Ze)( - e)}}{{4\pi {\varepsilon _0}r}} = - \dfrac{{Z{e^2}}}{{4\pi {\varepsilon _0}r}}$
Since it is a circular motion, this force will be the force required to keep the electron in circular motion and hence will be equal to the centripetal force.
$
\dfrac{{Z{e^2}}}{{4\pi {\varepsilon _0}{r^2}}} = \dfrac{{m{v^2}}}{r} \\
\Rightarrow r = \dfrac{{Z{e^2}}}{{4\pi {\varepsilon _0}m{v^2}}} \\ $
Now, using the 2nd postulate,
$
mvr = n\dfrac{h}{{2\pi }} \\
\Rightarrow v = \dfrac{{Z{e^2}}}{{2{\varepsilon _0}hn}} \\ $
Substituting the value of $v$ in $r$,
$
r = \dfrac{{Z{e^2}}}{{4\pi {\varepsilon _0}m{{\left( {\dfrac{{Z{e^{}}}}{{2{\varepsilon _0}hn}}} \right)}^2}}} \\
\Rightarrow r = \dfrac{{{\varepsilon _0}{h^2}{n^2}}}{{\pi mZ{e^2}}} \\ $
Now,the kinetic energy of the electron will be $K.E = \dfrac{1}{2}m{v^2} = \dfrac{{m{Z^2}{e^4}}}{{8{\varepsilon _0}^2{h^2}{n^2}}}$ and the potential energy will be $P.E = - \dfrac{{Z{e^2}}}{{4\pi {\varepsilon _0}r}} = - \dfrac{{m{Z^2}{e^4}}}{{4{\varepsilon _0}^2{h^2}{n^2}}}$

Therefore, the total energy of the electron will be $K.E + P.E = \dfrac{{m{Z^2}{e^4}}}{{8{\varepsilon _0}^2{h^2}{n^2}}} + \left( { - \dfrac{{m{Z^2}{e^4}}}{{4{\varepsilon _0}^2{h^2}{n^2}}}} \right)$
$T.E = - \dfrac{{m{Z^2}{e^4}}}{{8{\varepsilon _0}^2{h^2}{n^2}}}$
$\therefore {E_n} = - \dfrac{{m{Z^2}{e^4}}}{{8{\varepsilon _0}^2{h^2}{n^2}}}$
Here in the above equation, except $n$, all quantities are constant. Therefore, ${E_n} \propto \dfrac{1}{{{n^2}}}$.
Hence the expression for total energy of electron in ${n^{th}}$ Bohr orbit is $ - \dfrac{{m{Z^2}{e^4}}}{{8{\varepsilon _0}^2{h^2}{n^2}}}J$.

Note: The Bohr’s Atomic Model is only applicable for hydrogen like atoms having one electron. The 3rd Bohr’s postulate is “The electron can jump from one stationery orbit to the other by absorbing or emitting energy. If the electron jumps from higher to lower orbit, it will emit energy in the form of a photon. The energy of the photon is given by the difference between the higher and lower energy levels, ${E_2} - {E_1} = h\nu = \dfrac{{hc}}{\lambda }$ where $\lambda $ is the wavelength of the photon.