General guidance

Concepts and reason

The concept required to solve the given problem is Coulomb’s law and gravitational force.

Initially, calculate the gravitational force that acts on the electron.

Compute the electrostatic force between electron and proton with the help of Coulomb’s law and equate it to the gravitational force to compute the expression for separation between the electron and proton.

Finally, substitute the corresponding values into the expression thus obtained to compute the value of separation between electron and proton.

Fundamentals

Coulomb’s Law: The law states that the electrostatic force between any two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

It is given by,

${F_{{\rm{AB}}}} = \frac{{{k_{\rm{e}}}{q_{\rm{A}}}{q_{\rm{B}}}}}{{{r_{{\rm{BA}}}}^2}}$

Here, ${k_e}$ is the coulomb’s constant, ${q_{\rm{A}}}{\rm{ and }}{q_{\rm{B}}}$ are the charges and ${r_{{\rm{BA}}}}$ is the distance between two charges ${q_{\rm{A}}}{\rm{ and }}{q_{\rm{B}}}$ .

Gravitational force: It is the force with which earth attracts object towards itself. It is given by the formula,

$F = mg$

Here, $m$ is the mass and $g$ is the acceleration due to gravity.

First Step | All Steps | Answer Only

Step-by-step

Step 1 of 2

The gravitational force experienced by the electron will be given by,

${F_{\rm{e}}} = {m_{\rm{e}}}g$

Here, ${m_{\rm{e}}}$ is the mass of electron and $g$ is the acceleration due to gravity.

Substitute $9.1 \times {10^{ - 31}}{\rm{ kg}}$ for ${m_e}$ and $9.8{\rm{ m / }}{{\rm{s}}^2}$ for $g$ in the above equation.

$\begin{array}{c}\\{F_{\rm{e}}} = \left( {9.1 \times {{10}^{ - 31}}{\rm{ kg}}} \right)\left( {9.8{\rm{ m / }}{{\rm{s}}^2}} \right)\\\\ = 89.2 \times {10^{ - 31}}{\rm{ N}}\\\end{array}$

Explanation | Common mistakes | Hint for next step

The gravitational force acting on the electron was calculated by the formula, $F = mg$ .

Step 2 of 2

The electrostatic force that acts between the electron and the proton will be,

$F = \frac{{{k_{\rm{e}}}{q_1}{q_2}}}{{{r^2}}}$

Here, ${k_{\rm{e}}}$ is the Coulomb’s constant, ${q_1}{\rm{ and }}{q_2}$ are the charges and $r$ is the distance between the charges.

For the electrostatic force to be equal to the gravitational force that acts on the electron equate $F{\rm{ and }}{F_{\rm{e}}}$ .

$F = {F_{\rm{e}}}$

Substitute $89.2 \times {10^{ - 31}}{\rm{ N}}$ for ${F_{\rm{e}}}$ and $\frac{{{k_{\rm{e}}}{q_1}{q_2}}}{{{r^2}}}$ for $F$ in the above equation.

$\frac{{{k_{\rm{e}}}{q_1}{q_2}}}{{{r^2}}} = 89.2 \times {10^{ - 31}}{\rm{ N}}$

Rearrange the above equation.

$r = \sqrt {\frac{{{k_{\rm{e}}}{q_1}{q_2}}}{{89.2 \times {{10}^{ - 31}}{\rm{ N}}}}}$

Substitute $9 \times {10^9}{\rm{ N}} \cdot {{\rm{m}}^2}{\rm{ / }}{{\rm{C}}^2}$ for ${k_e}$ and $1.6 \times {10^{ - 19}}{\rm{ C}}$ for ${q_1}{\rm{ and }}{q_2}$ in the above equation.

$\begin{array}{c}\\r = \sqrt {\frac{{\left( {9 \times {{10}^9}{\rm{ N}} \cdot {{\rm{m}}^2}{\rm{ / }}{{\rm{C}}^2}} \right)\left( {1.6 \times {{10}^{ - 19}}{\rm{ C}}} \right)\left( {1.6 \times {{10}^{ - 19}}{\rm{ C}}} \right)}}{{89.2 \times {{10}^{ - 31}}{\rm{ N}}}}} \\\\ = 5.08{\rm{ m}}\\\end{array}$

Thus, the separation between the electron and proton should be $5.08{\rm{ m}}$ which is not possible since the maximum distance between the electron and proton can be ${10^{ - 11}}{\rm{ m}}$ that is of the order of radius of atom.

The proton must be at a distance of $5.08{\rm{ m}}$ from the electron.

Explanation | Common mistakes

The electrostatic force between the electron and proton was calculated by the formula, $F = \frac{{{k_{\rm{e}}}{q_1}{q_2}}}{{{r^2}}}$ and equated to the gravitational force acting on the electron to determine the separation between the electron and proton required to produce the electrostatic force equal to the gravitational force.

Answer

The proton must be at a distance of $5.08{\rm{ m}}$ from the electron.

An electron at Earth's surface experiences a gravitational force meg. How far away can a proton be and still produce the same force on the electron? (Your answer should show why gravity is unimportant on the molecular scale!)

Public Answer