Question

In the ground state of the hydrogen atom, the electron has a total energy of -13.06 eV. What are (a) its kinetic energy and (b) its potential energy if the electron is one Bohr radius from the central nucleus?

Short answer

The recoilingspeed of the hydrogen atom is 4.08 m/s.

Step-by-step solution

Describe the expression for potential energy and kinetic energy:

The expression for the potential energy of the electron is given by,

$\mathbf{U}\mathbf{=}\mathbf{-}\mathbf{k}\frac{{\mathbf{e}}^{\mathbf{2}}}{\mathbf{a}}$ ….. (1)

Here, $\mathbf{a}\mathbf{=}\mathbf{52}\mathbf{.}\mathbf{292}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{12}}\mathbf{}\mathbf{m}$is the Bohr radius,$\mathit{k}\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{99}\mathbf{\times }{\mathbf{10}}^{\mathbf{9}}\mathbf{N}\mathbf{.}{\mathbf{m}}^{\mathbf{2}}\mathbf{/}{\mathbf{C}}^{\mathbf{2}}$is the Coulomb’s constant, and $\mathbf{e}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{60}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{19}}\mathbf{}\mathbf{C}$is charge of the electron.

The expression for kinetic energy is given by,

$\mathbf{K}\mathbf{=}\mathbf{E}\mathbf{-}\mathbf{U}$ ….. (2)

Here, E is the total energy, K is the kinetic energy and U is the potential energy..

(a) Define the kinetic energy of the electron:

The kinetic energy of the electron can be calculated as follows.

$$\begin{gathered} \mathrm{K}=\mathrm{E}-\mathrm{U} \\ =\mathrm{E}-\frac{{\mathrm{ke}}^2}{\mathrm{a}} \end{gathered}$$

$$\begin{gathered} \mathrm{K}=\left(-13.6\mathrm{eV}\right)-\left(-8.99\times {10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2\times \frac{{\left(1.6\times {10}^{-19}\mathrm{C}\right)}^2}{\left(5.292\times {10}^{-11}\mathrm{m}\right)}\right) \\ =\left(-13.6\mathrm{eV}\right)-\left(-4.36\times {10}^{-18}\mathrm{J}\times \frac{1\mathrm{eV}}{1.6\times {10}^{-19}\mathrm{J}}\right) \\ =\left(-13.6\mathrm{eV}\right)+27.25\mathrm{eV} \\ =13.6\mathrm{eV} \end{gathered}$$

Therefore, the kinetic energy of the electron is 13.6 eV.

(b) Define the potential energy of the electron:

The potential energy of the electron can be calculated as follows.

$$\begin{gathered} \mathrm{K}=-\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{{\mathrm{e}}^2}{\mathrm{a}} \\ =-8.99\times {10}^9\mathrm{N}.{\mathrm{m}}^2/{\mathrm{C}}^2\times \frac{{\left(1.6\times {10}^{-19}\mathrm{C}\right)}^2}{\left(5.292\times {10}^{-11}\mathrm{m}\right)} \\ =-4.36\times {10}^{-18}\mathrm{J}\times \frac{1\mathrm{eV}}{1.6\times {10}^{-19}\mathrm{J}} \\ =-27.2\mathrm{eV} \end{gathered}$$

Hence, the potential energy of the electron is -27.2 eV.