(III) In the Bohr model of the hydrogen atom, an electron orbits a proton (the nucleus) in a circular orbit of radius \({\bf{0}}{\bf{.53 \times 1}}{{\bf{0}}^{{\bf{ - 10}}}}\;{\bf{m}}\). (a) What is the electric potential at the electron’s orbit due to the proton? (b) What is the kinetic energy of the electron? (c) What is the total energy of the electron in its orbit? (d) What is the ionization energy— that is, the energy required to remove the electron from the atom and take it to \({\bf{r = }}\infty \), at rest? Express the results of parts (b), (c), and (d) in joules and eV.

The electric potential due to a point charge is given as:

\(V = k\frac{Q}{r}\) … (i)

Here, k is the Coulomb’s constant, Q is the charge and r is the distance from the charge.

Near a positive charge, the potential has a large positive value and decreases to zero at large distances.

Near a negative charge, the potential has a negative value and increases towards zero at large distances.

The radius of orbit is\(r = 0.53 \times {10^{ - 10}}\;{\rm{m}}\).

Charge on the proton is, \(q = 1.6 \times {10^{ - 19}}\;{\rm{C}}\)

From equation (i), the electric potentialdue to proton is given as:

\(V = \frac{{kq}}{r}\)

Substitute the values in the above expression.

\(\begin{aligned}V &= \left( {\frac{{\left( {9 \times {{10}^9}\;{\rm{N}} \cdot {{\rm{m}}^2}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right)\left( {1.6 \times {{10}^{ - 19}}\;{\rm{C}}} \right)}}{{\left( {0.53 \times {{10}^{ - 10}}\;{\rm{m}}} \right)}}} \right)\\V &\approx 27\;{\rm{V}}\end{aligned}\)

Thus, the electric potential at the electron’s orbit due to proton is \(27\;{\rm{V}}\).

The relation of kinetic energyis given by,

\(KE = \frac{1}{2}k\frac{{{q^2}}}{r}\)

Substitute the values in the above expression.

\(\begin{aligned}KE &= \frac{1}{2}\left( {9.0 \times {{10}^9}\;{\rm{N}} \cdot {{\rm{m}}^2}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right)\frac{{{{\left( {1.6 \times {{10}^{ - 19}}\;{\rm{C}}} \right)}^2}}}{{\left( {0.53 \times {{10}^{ - 10}}\;{\rm{m}}} \right)}}\\KE &= 2.17 \times {10^{ - 18}}\;{\rm{J}}\end{aligned}\)

The kinetic energy in terms of electron volts is,

\(\begin{aligned}KE &= \left( {2.17 \times {{10}^{ - 18}}\;{\rm{J}} \times \frac{{1\;{\rm{eV}}}}{{1.6 \times {{10}^{ - 19}}\;{\rm{J}}}}} \right)\\KE &= 13.5\;{\rm{eV}}\end{aligned}\)

Thus,the kinetic energy of the electron is \(2.17 \times {10^{ - 18}}\;{\rm{J}}\)and \(13.5\;{\rm{eV}}\).

The relation of total energy is given by,

\(\begin{aligned}E &= PE + KE\\E &= - qV + \frac{1}{2}k\frac{{{q^2}}}{r}\\E &= - k\frac{{{q^2}}}{r} + \frac{1}{2}k\frac{{{q^2}}}{r}\\E &= - \frac{1}{2}k\frac{{{q^2}}}{r}\end{aligned}\)

Substitute the values in the above expression.

\(\begin{aligned}KE &= - \frac{1}{2}\left( {9.0 \times {{10}^9}\;{\rm{N}} \cdot {{\rm{m}}^2}{\rm{/}}{{\rm{C}}^{\rm{2}}}} \right)\frac{{{{\left( {1.6 \times {{10}^{ - 19}}\;{\rm{C}}} \right)}^2}}}{{\left( {0.53 \times {{10}^{ - 10}}\;{\rm{m}}} \right)}}\\KE &= - 2.17 \times {10^{ - 18}}\;{\rm{J}}\end{aligned}\)

The kinetic energy in terms of electron volts is,

\(\begin{aligned}KE &= - 2.17 \times {10^{ - 18}}\;{\rm{J}} \times \frac{{1\;{\rm{eV}}}}{{1.6 \times {{10}^{ - 19}}\;{\rm{J}}}}\\KE &= - 13.5\;{\rm{eV}}\end{aligned}\)

Thus, the Total energy of the electron is \( - 2.17 \times {10^{ - 18}}\;{\rm{J}}\)and \( - 13.5\;{\rm{eV}}\).

When the electron is taken to infinite at rest, the potential and kinetic energy will be zero. So, the energy required for the electron to have total energy equal to zero is \(2.17 \times {10^{ - 18}}\;{\rm{J}}\)or\(13.5\;{\rm{eV}}\).

Thus, the ionization energy is \(2.17 \times {10^{ - 18}}\;{\rm{J}}\)or \(13.5\;{\rm{eV}}\).