Question

In a simple model of the hydrogen atom, the electron revolves in a circular orbit around the proton with a speed of \({\bf{2}}{\bf{.2 \times 1}}{{\bf{0}}^{\bf{6}}}\;{\bf{m/s}}\). Determine the radius of the electron’s orbit. (Hint: See Chapter 5 on circular motion.)

Short answer

The radius of the electron’s orbit is \(5.21 \times {10^{ - 11}}\,{\rm{m}}\).

Step-by-step solution

Understanding the electric force

The proton in the nucleus of the hydrogen atom exerts an electric force on the electron revolving in a circular orbit.

The expression for the electric force is given as:

\(F = k\frac{{{q_1}{q_2}}}{{{r^2}}}\) … (i)

Here, k is Coulomb’s constant,\({q_1},\;{q_2}\)are the charges on electron and proton and r is the separation between the charges.

Given Data

The speed of the electron is, \(v = 2.2 \times {10^6}\;{\rm{m/s}}\).

Determination of radius of the electron’s orbit

In the circular motion, the electric force will be equal to the centripetal force.

\(\begin{aligned}{c}F &= k\frac{{{q^2}}}{{{r^2}}}\\\frac{{m{v^2}}}{r} &= k\frac{{{q^2}}}{{{r^2}}}\\r &= k\frac{{{q^2}}}{{m{v^2}}}\end{aligned}\)

Here, m is the mass of the electron, q is the charge and r is the radius.

Substitute the values in the above expression.

\(\begin{aligned}{l}r &= \left( {\left( {9.0 \times {{10}^9}\;{\rm{N}} \cdot {{\rm{m}}^2}{\rm{/}}{{\rm{C}}^2}} \right)\frac{{{{\left( {1.6 \times {{10}^{ - 19}}\;{\rm{C}}} \right)}^2}}}{{\left( {9.1 \times {{10}^{ - 31}}\;{\rm{kg}}} \right){{\left( {2.2 \times {{10}^6}\;{\rm{m/s}}} \right)}^2}}}} \right)\\r &= 5.21 \times {10^{ - 11}}\,{\rm{m}}\end{aligned}\)

Thus, the radius of the electron’s orbit is \(5.21 \times {10^{ - 11}}\,{\rm{m}}\).