Chapter 4, Problem 16P

To determine

The speed and radial acceleration for a ground-state electron in the hydrogen atom.

Answer to Problem 16P

The speed for a ground-state electron in the hydrogen atom is $2.19\times {10}^6\text{m/s}$ and radial acceleration is $9.07\times {10}^{22}{\text{m/s}}^{\text{2}}$ while for lithium atom speed is $6.57\times {10}^6\text{m/s}$ and acceleration is $2.45\times {10}^{24}{\text{m/s}}^{\text{2}}$.

Explanation of Solution

If an electron rotates around the nucleus then the expression for speed is given below:

    $v=\frac{e}{\sqrt{4\pi {\epsilon }_{0}mr}}$

Here, $v$ is the speed of an electron, $e$ is the charge on electron, $r$ is the radius and $m$ is the mass of electron.

Multiply by $c$ in both numerator and denominator.

    $v=\frac{ec}{\sqrt{4\pi {\epsilon }_0}\sqrt{m{c}^{2}r}}$        (I)

Value of $m{c}^2$ in electron Volt is:

    $\begin{array}{c}m{c}^2=\frac{9.1\times {10}^{-31}\times {\left(3\times {10}^8\right)}^2}{1.6\times {10}^{-19}}\text{eV}\\ =\text{511875}\text{eV}\end{array}$ 

Write the expression for radial acceleration.

    $a=\frac{v^2}{r}$        (II)

Here, $a$ is the radial acceleration.

For hydrogen like lithion ion ${\text{Li}}^{++}$.

Write the expression for maximum Coulomb force.

    $F=\frac{Z{e}^2}{4\pi {\epsilon }_0{r}^2}$        (III)

Here, $F$ is the Coulomb force, $Z$ is the atomic number of the metal, $r$ is the radius and $e$ is the charge of electron.

Write the expression for centripetal force.

    $F=\frac{m{v}^2}{r}$        (IV)

Here, $F$ is the centripetal force.

From equation (III) and (IV) solve and rearrange the equation for value of $v^2$.

    $v^2=\frac{Z{e}^2}{4\pi {\epsilon }_{0}rm}$        (V)

Write the expression for radius.

    $r=\frac{4\pi {\epsilon }_0{h}^2}{Zm{e}^2}=\frac{a_0}{Z}$

Here, $a_0$ is the Bohr’s radius.

Substitute $\frac{a_0}{Z}$ for $r$ in equation (V).

    $v^2=\frac{Z^2{e}^2}{4\pi {\epsilon }_0{a}_{0}m}$        (VI)

Conclusion:

Substitute $\text{511875}\text{eV}$ for $m{c}^2$, $9\times {10}^9$ for $\frac{1}{4\pi {\epsilon }_0}$ , $1.6\times {10}^{-19}$ for $e$  and $5.29\times {10}^{-11}\text{m}$ for $r$ in equation (I).

    $\begin{array}{c}v=\frac{\sqrt{\left(9\times {10}^9\times 1.6\times {10}^{-19}\text{eV}\text{.m}\right)}c}{\sqrt{\left(\text{511875}\text{eV}\right)\left(5.29\times {10}^{-11}\text{m}\right)}}\\ =\frac{\sqrt{1.44\text{eV}\cdot \text{nm}}\left(3\times {10}^8\right)}{\sqrt{\left(\text{511875}\text{eV}\right)\left(0.0529\text{nm}\right)}}\\ =2.19\times {10}^6\text{m/s}\end{array}$

Substitute $2.19\times {10}^6\text{m/s}$ for $v$ and $5.29\times {10}^{-11}\text{m}$ for $r$ in equation (II).

    $\begin{array}{c}a=\frac{{\left(2.19\times {10}^6\text{m/s}\right)}^2}{\left(5.29\times {10}^{-11}\text{nm}\right)}\\ =9.07\times {10}^{22}{\text{m/s}}^{\text{2}}\end{array}$

For Lithium ion:

Substitute $\text{511875}\text{eV/}{c}^2$ for $m$, $3$ for $Z$ , $9\times {10}^9$ for $\frac{1}{4\pi {\epsilon }_0}$ , $1.6\times {10}^{-19}$ for $e$  and $5.29\times {10}^{-11}\text{m}$ for $a_0$ in equation (VI).

    $\begin{array}{c}{v}^2=\frac{{\left(3\right)}^2\left(1.6\times {10}^{-19}eV\right)\left(9\times {10}^9\right)}{\left(5.29\times {10}^{-11}\text{m}\right)\left(\text{511875}\text{eV/}{c}^2\right)}\\ v^2=4.78\times {10}^{-4}{c}^2\\ v=2.19\times {10}^{-2}c\\ =6.57\times {10}^6\text{m/s}\end{array}$

Substitute $6.57\times {10}^6\text{m/s}$ for $v$ and $5.29\times {10}^{-11}\text{m}$ for $r$ in equation (II).

    $\begin{array}{c}a=\frac{{\left(6.57\times {10}^6\text{m/s}\right)}^2}{\left(5.29\times {10}^{-11}\text{nm}\right)/3}\\ =2.45\times {10}^{24}{\text{m/s}}^{\text{2}}\end{array}$

Thus, the speed for a ground-state electron in the hydrogen atom is $2.19\times {10}^6\text{m/s}$ and radial acceleration is $9.07\times {10}^{22}{\text{m/s}}^{\text{2}}$ while for lithium atom speed is $6.57\times {10}^6\text{m/s}$ and acceleration is $2.45\times {10}^{24}{\text{m/s}}^{\text{2}}$.