Question

Using Rutherford’s model of atom, derive the expression for the total energy of the electron in a hydrogen atom. What is the significance of the total negative energy possessed by the electron?
OR
Using Bohr’s postulates of the atomic model derive the expression for the radius of $ nth $ electron orbit. Hence obtain the expression for Bohr radius.

Answer 1

Hint :In order to solve this question, we are going to first use the equation for Rutherford's model of an atom, and find the value of the kinetic energy from it. After that the total energy is found from the kinetic and potential energies. In the second part, applying the Bohr’s atomic model, the radius is found by using the centripetal force.
According to the Rutherford’s model of an atom
 $ \dfrac{{m{v^2}}}{r} = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{Z{e^2}}}{{{r^2}}} $
 $ T.E. = P.E. + K.E. $ , $ P.E. = - \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{Z{e^2}}}{r} $
According to the Bohr’s postulate, we have
 $ mvr = \dfrac{{nh}}{{2\pi }} $

Complete Step By Step Answer:
According to the Rutherford’s model of an atom, we have
 $ \dfrac{{m{v^2}}}{r} = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{Z{e^2}}}{{{r^2}}} $
Now rewriting this equation, in order to find the kinetic energy, we get
 $ m{v^2} = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{Z{e^2}}}{r} \\
   \Rightarrow \dfrac{1}{2}m{v^2} = \dfrac{1}{{8\pi {\varepsilon _0}}}\dfrac{{Z{e^2}}}{r} \\ $
The total energy equals the sum of the potential energy and the kinetic energy
 $ T.E. = P.E. + K.E. $
Now the potential energy, is given by
 $ P.E. = - \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{Z{e^2}}}{r} \\
  K.E. = \dfrac{1}{{8\pi {\varepsilon _0}}}\dfrac{{Z{e^2}}}{r} \\ $
Therefore, the total energy becomes
 $ T.E. = - \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{Z{e^2}}}{r} + \dfrac{1}{{8\pi {\varepsilon _0}}}\dfrac{{Z{e^2}}}{r} \\
   \Rightarrow T.E. = - \dfrac{1}{{8\pi {\varepsilon _0}}}\dfrac{{Z{e^2}}}{r} \\ $
The negative sign signifies that the electron – nucleus pair is a bound or attractive system.
 OR
According to the Bohr’s model of atom, the electrons revolve around the nucleus. Only those orbits are allowed for which the angular momentum is an integral multiple of $ \dfrac{h}{{2\pi }} $
So, according to the Bohr’s postulate, we have
 $ mvr = \dfrac{{nh}}{{2\pi }} $
Now the centripetal force is given by
 $ \dfrac{{m{v^2}}}{r} = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{Z{e^2}}}{{{r^2}}} - - - \left( 1 \right) $
Squaring both sides of the equation $ mvr = \dfrac{{nh}}{{2\pi }} $
 $ {m^2}{v^2}{r^2} = \dfrac{{{n^2}{h^2}}}{{4{\pi ^2}}} - - - \left( 2 \right) $
Now, solving $ \left( 1 \right) $ and $ \left( 2 \right) $
 $ \dfrac{{{n^2}{h^2}}}{{4{\pi ^2}}} = \dfrac{1}{{4\pi {\varepsilon _0}}}Zm{e^2}r \\
   \Rightarrow r = \dfrac{{{\varepsilon _0}{n^2}{h^2}}}{{\pi Z{e^2}m}} \\ $
Thus, this is the required expression for the Bohr’s radius.

Note :
It may be noted that in both the parts, we have used the expression for the centripetal force of the electrons revolving around the nucleus because it is the main driving force in a circular motion. Additionally, the particular equations that were given by both Rutherford and Bohr respectively have been used to obtain the required result.