Question

Prove that energy of ${{n}^{th}}$ orbit is inversely proportional to square of the principal quantum number.

Answer 1

Hint:We know that an electron is revolving around the nucleus in a uniform circular motion, and for a body to move in circular motion there must be a centripetal force acting towards the center. In case of an electron of mass $m$ revolving in its orbit of radius $r$ this centripetal force is provided by the electrostatic force of attraction between the electron and nucleus.
$\dfrac{m{{v}^{2}}}{r}=\dfrac{1}{4\pi {{\in }_{{}^\circ }}}.\dfrac{{{e}^{2}}}{{{r}^{2}}}$
From above find the kinetic energy of the electron.Also, find the potential energy of the electron by: $PE=qV\text{ where, }V=\dfrac{q}{4\pi {{\varepsilon }_{{}^\circ }}r}$
The total energy of the electron will be given by the sum of both its kinetic energy and potential energy.

Complete step by step answer:
1. We will consider an electron revolving in the ${{n}^{th}}$ orbit around the hydrogen nucleus. Let m be the mass, -e be the charge of electron, v be the linear speed of electron, r be the radius of ${{n}^{th}}$ orbit. According to Bohr’s first postulate, the centripetal force acting on the electron = the electrostatic force of attraction exerted by the nucleus on the electron. Hence,
$\dfrac{m{{v}^{2}}}{r}=\dfrac{1}{4\pi {{\in }_{{}^\circ }}}.\dfrac{{{e}^{2}}}{{{r}^{2}}}$
  $\Rightarrow m{{v}^{2}}=\dfrac{1}{4\pi {{\in }_{{}^\circ }}}.\dfrac{{{e}^{2}}}{r}$..........................………....eq1
where ${{\in }_{{}^\circ }}$ = permittivity of the free space.
2. Kinetic energy of electron is given by :
$KE=\dfrac{m{{v}^{2}}}{2}$ .
Substituting eq1 here, we get:
$\dfrac{m{{v}^{2}}}{2}=\dfrac{{{e}^{2}}}{2\times 4\pi {{\in }_{{}^\circ }}r}=\dfrac{{{e}^{2}}}{8\pi {{\in }_{{}^\circ }}r}$ .......…………….eq2
3. Potential energy of electron is given by:
$PE=qV=(-e).\dfrac{e}{4\pi {{\in }_{{}^\circ }}r}$ $=\dfrac{-{{e}^{2}}}{4\pi {{\in }_{{}^\circ }}r}$ ………………eq3
4. Now, total energy at ${{n}^{th}}$ orbit is: E = KE + PE.
So, from equations 2 and 3 we have,
$E=\left( \dfrac{-{{e}^{2}}}{4\pi {{\in }_{{}^\circ }}r} \right)+\left( \dfrac{{{e}^{2}}}{8\pi {{\in }_{{}^\circ }}r} \right)$
$\Rightarrow E=\left( \dfrac{{{e}^{2}}-2{{e}^{2}}}{8\pi {{\in }_{{}^\circ }}r} \right)$
$\Rightarrow E=\left( \dfrac{-{{e}^{2}}}{8\pi {{\in }_{{}^\circ }}r} \right)$ ………………eq4
5. But $r=\dfrac{{{\in }_{{}^\circ }}{{h}^{2}}}{\pi m{{e}^{2}}}.{{n}^{2}}$ (h=Planck’s constant).
Substituting this value in equation 4, we get:
\[E=-\dfrac{1}{8\pi {{\in }_{{}^\circ }}}\dfrac{{{e}^{2}}}{\dfrac{{{\in }_{{}^\circ }}{{h}^{2}}}{\pi m{{e}^{2}}}.{{n}^{2}}}\]
\[E=-\dfrac{{{e}^{2}}}{8\pi {{\in }_{{}^\circ }}}\left( \dfrac{\pi m{{e}^{2}}}{{{\in }_{{}^\circ }}{{h}^{2}}{{n}^{2}}} \right)=-\dfrac{m{{e}^{4}}}{8{{\in }^{2}}_{{}^\circ }{{h}^{2}}{{n}^{2}}}\] ………….eq5
This is the expression for energy of an electron in an orbit of Bohr’s hydrogen atom. Now, as we know in the above expression, m, e, h and ${{\in }_{{}^\circ }}$ are constant, i.e.
\[\dfrac{m{{e}^{4}}}{8{{\in }^{2}}_{{}^\circ }{{h}^{2}}}\] = constant, we get:
$Energy=const.\times \dfrac{1}{{{n}^{2}}} \\
\therefore E\propto \dfrac{1}{{{n}^{2}}} \\$
Hence proved that energy of an electron is inversely proportional to square of principle quantum number $n$.

Note:The negative sign in equation: \[E=-\dfrac{m{{e}^{4}}}{8{{\in }^{2}}_{{}^\circ }{{h}^{2}}{{n}^{2}}}\]
shows that the electron is bound to the nucleus by an attractive force and hence energy must be supplied to the electron to make it free from influence of the nucleus. We need to be careful while solving such problems that all the quantities are to be taken in standard SI unless specified. The negative sign of total energy signifies that the orbit is bounded.