Chapter 16, Problem 85QAP

To determine

(a)

The magnitude and direction of the electric field just outside surface of nucleus.

Answer to Problem 85QAP

The magnitude of the electric field just outside surface of nucleus is, $E=1.769\times {10}^{21}\text{N/C}$ and it directed outwards from the nucleus.

Explanation of Solution

Given:

Diameter of atom, $D=1.0\times {10}^{-10}\text{m}$

Radius of atom, $R=0.5\times {10}^{-10}\text{m}$

Diameter of nucleus, $d=9.2\times {10}^{-16}\text{m}$

Radius of nucleus, $r=4.6\times {10}^{-16}\text{m}$

Charge on electron, $q=1.6\times {10}^{-19}\text{C}$

Mass of electron, $m=9.1\times {10}^{-31}\text{kg}$

Formula used:

The electric field is given by,
  $E=\frac{kq}{r^2}$

Where,
  $E$ =Electric field
  $q$ = Electric charges
  $r$ =Distance from charge

Calculation:

The electric field is given by,
  $\begin{array}{l}E=\frac{kq}{r^2}\\ E=\frac{9\times {10}^9\times 26\times 1.6\times {10}^{-19}}{4.6\times 4.6\times {10}^{-30}}\\ E=1.769\times {10}^{21}\text{N/C}\end{array}$

The electric field is directed outwards from the nucleus because nucleus acts as point source.

To determine

(b)

The magnitude and direction of the electric field at the distance of outermost electron.

Answer to Problem 85QAP

The magnitude and direction of the electric field at the distance of outermost electron is, $E=1.497\times {10}^{13}\text{N/C}$

Explanation of Solution

Given:

Diameter of atom, $D=1.0\times {10}^{-10}\text{m}$

Radius of atom, $R=0.5\times {10}^{-10}\text{m}$

Diameter of nucleus, $d=9.2\times {10}^{-15}\text{m}$

Radius of nucleus, $r=4.6\times {10}^{-15}\text{m}$

Charge on electron, $q=1.6\times {10}^{-19}\text{C}$

Mass of electron, $m=9.1\times {10}^{-31}\text{kg}$

Formula used:

The electric field is given by,
  $E=\frac{kq}{r^2}$

Where,
  $E$ =Electric field
  $q$ = Electric charges
  $r$ =Distance from charge

Calculation:

The electric field is given by,
  $\begin{array}{l}E=\frac{kq}{R^2}\\ E=\frac{9\times {10}^9\times 26\times 1.6\times {10}^{-19}}{0.5\times 0.5\times {10}^{-20}}\\ E=1.497\times {10}^{13}\text{N/C}\end{array}$

The electric field is directed outwards from the nucleus.

To determine

(c)

The magnitude and direction of the acceleration of the outermost electron.

Answer to Problem 85QAP

The magnitude and direction of the acceleration of outermost electron is, $a=2.632\times {10}^{24}{\text{m/s}}^{\text{2}}$ and it is directed outwards from the nucleus.

Explanation of Solution

Given:

Diameter of atom, $D=1.0\times {10}^{-10}\text{m}$

Radius of atom, $R=0.5\times {10}^{-10}\text{m}$

Diameter of nucleus, $d=9.2\times {10}^{-15}\text{m}$

Radius of nucleus, $r=4.6\times {10}^{-15}\text{m}$

Charge on electron, $q=1.6\times {10}^{-19}\text{C}$

Mass of electron, $m=9.1\times {10}^{-31}\text{kg}$

Electric field, $E=1.497\times {10}^{13}\text{N/C}$

Formula used:

The force acting on the ball is,
$F=qE$

Where,
  $F$ =Electrostatic force
  $q$ = Electric charge
  $E$ =Electric field

Calculation:

The force is given by,
  $F=qE\mathrm{...}(1)$

But we know that,
$F=ma\mathrm{...}(2)$

From equation (1) and (2)
$\begin{array}{l}qE=ma\\ a=\frac{qE}{m}\\ a=\frac{1.6\times {10}^{-19}\times 1.497\times {10}^{13}}{9.10\times {10}^{-31}}\\ a=2.632\times {10}^{24}{\text{m/s}}^{\text{2}}\end{array}$

The acceleration is directed outwards from the nucleus.