Question

If a proton has a radius R and the charge was uniformly distributed, calculate using Bohr theory, the ground state energy of a H-atom when (i) R=\[0.1\mathop A\limits^ \circ \], and (ii) R=\[10\mathop A\limits^ \circ \]

Answer 1

Hint: In Hydrogen atoms there is one proton in the nucleus and one electron revolves around the nucleus. The centripetal columbic force balanced the centrifugal force which acts outward.

Formula used:
\[K = \dfrac{{m{v^2}}}{2}\]
where K is the kinetic energy of a particle of mass m moving with speed v.
\[U = \dfrac{{K{q_1}{q_2}}}{r}\]
where U is the electric potential energy of the system of two charges \[{q_1}\] and \[{q_2}\] kept at the separation of r.

Complete step by step solution:
(i) Consider the nucleus of the H-atom as a point positive charge and when the electron moves around it then there is Coulomb force between them which balances the outward centrifugal force. If the speed of the electron is v then,
\[\dfrac{{{m_e}{v^2}}}{R} = \dfrac{{K{e^2}}}{{{R^2}}} \ldots \left( 1 \right)\]

According to the Bohr's postulate, angular momentum of the electron in Hydrogen atom in nth principle quantum number is given as,
\[L = \dfrac{{nh}}{{2\pi }}\]
When electron is moving in the same energy level with linear speed v, then the angular momentum will be,
\[{m_e}vR = \dfrac{{nh}}{{2\pi }} \\ \]
\[\Rightarrow {v^2} = {\left( {\dfrac{{nh}}{{2\pi {m_e}R}}} \right)^2} \\ \]
\[\Rightarrow \dfrac{{{m_e}{v^2}}}{R} = \dfrac{{{n^2}{h^2}}}{{4{\pi ^2}{m_e}{R^3}}} \ldots \left( 2 \right)\]
The radius of the atom is given as \[R = 0.1\mathop A\limits^ \circ \]

The total energy is the sum of the kinetic energy and the potential energy.
\[E = K + U\]
\[\Rightarrow E = \dfrac{{{m_e}{v^2}}}{2} + \dfrac{{K \times \left( { - e} \right) \times \left( e \right)}}{R} \\ \]
\[\Rightarrow E = \dfrac{{K{e^2}}}{{2R}} - \dfrac{{K{e^2}}}{R}\] from first equation,
\[\Rightarrow E = - \dfrac{{K{e^2}}}{{2R}}\]

Putting the values, we get
\[E = - \dfrac{{9 \times {{10}^9} \times {{\left( {1.6 \times {{10}^{ - 19}}} \right)}^2}}}{{2 \times 0.1 \times {{10}^{ - 10}}}}J \\ \]
\[\Rightarrow E = - 1.15 \times {10^{ - 17}}J \\ \]
\[\Rightarrow E = - \dfrac{{1.15 \times {{10}^{ - 17}}}}{{1.6 \times {{10}^{ - 19}}}}eV \\ \]
\[\therefore E = - 72\,eV\]

Therefore, the energy of the electron is -72 eV.

(ii) The radius of the atom is given as \[R = 10\mathop A\limits^ \circ \]. The total energy is the sum of the kinetic energy and the potential energy.
\[E = K + U\]
\[\Rightarrow E = \dfrac{{{m_e}{v^2}}}{2} + \dfrac{{K \times \left( { - e} \right) \times \left( e \right)}}{R} \\ \]
\[\Rightarrow E = \dfrac{{K{e^2}}}{{2R}} - \dfrac{{K{e^2}}}{R}\] from first equation,
\[\Rightarrow E = - \dfrac{{K{e^2}}}{{2R}}\]

Putting the values, we get
\[E = - \dfrac{{9 \times {{10}^9} \times {{\left( {1.6 \times {{10}^{ - 19}}} \right)}^2}}}{{2 \times 10 \times {{10}^{ - 10}}}}J \\ \]
\[\Rightarrow E = - 1.15 \times {10^{ - 18}}J \\ \]
\[\Rightarrow E = - \dfrac{{1.15 \times {{10}^{ - 18}}}}{{1.6 \times {{10}^{ - 19}}}}eV \\ \]
\[\therefore E = - 7.2\,eV\]

Therefore, the energy of the electron is -7.2 eV.

Note: We should be careful while writing the electric potential energy. Here we take the charge with the sign of the charge. The final energy of the electron is calculated in electron-volt units.