Answer

Verified

310.2k+ views

**Hint:**Before we approach the solution, it is important to understand the nature of force acting among the electrons and protons. Since they are charged bodies, there is an electrostatic force acting between them, which is given by Coulomb's law. Also, weight is equal to the mass times the acceleration due to gravity.

**Complete step by step answer:**

The problem has two parts: First part is if the charge is electron and second part is if the charge is a proton.

**Part A: Electrons**

Consider two electrons placed at a distance r. The electrostatic force acting between them is given by Coulomb's law which states the force between the two charges is directly proportional to the product of the charges and inversely proportional to the distance between the charges.

$F = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{q_1}{q_2}}}{{{r^2}}}$

where ${\varepsilon _0}$ = absolute permittivity of free space.

The charge on an electron is equal to –

$e = 1. 602 \times {10^{ - 19}}C$

The weight of the electron is equal to the product of mass and acceleration due to gravity.

$W = mg$

Mass of electron, $m = 9. 1 \times {10^{ - 31}}kg$

Acceleration due to gravity, $g = 10m{s^{ - 2}}$

Weight, $W = 9. 1 \times {10^{ - 31}} \times 10 = 9. 1 \times {10^{ - 30}}N$

Given that the force acting between the electrons is equal to the weight of an electron, we have –

$F = W$

$ \Rightarrow \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{q_1}{q_2}}}{{{r^2}}} = W$

Substituting, we have –

$9 \times {10^9} \times \dfrac{{{{\left( {1. 6 \times {{10}^{ - 19}}} \right)}^2}}}{{{r^2}}} = 9. 1 \times {10^{ - 30}}$

$ \Rightarrow {r^2} = \dfrac{{{{\left( {1. 6 \times {{10}^{ - 19}}} \right)}^2} \times 9 \times {{10}^9}}}{{9. 1 \times {{10}^{ - 30}}}}$

$ \Rightarrow {r^2} = \dfrac{{2. 56 \times {{10}^{ - 38}} \times 9 \times {{10}^9}}}{{9. 1 \times {{10}^{ - 30}}}}$

\[ \Rightarrow {r^2} = \dfrac{{2. 56 \times 9}}{{9. 1}} \times {10^{ - 38 + 9 + 30}}\]

\[ \Rightarrow {r^2} = 2. 5318 \times 10 = 25. 318\]

\[\therefore r = \sqrt {25. 318} = 5. 031m\]

Thus, the distance of separation is equal to 5.03 m

**Part B: Protons**

Consider two protons placed at a distance r. The electrostatic force acting between them is given by Coulomb's law which states the force between the two charges is directly proportional to the product of the charges and inversely proportional to the distance between the charges.

$F = \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{q_1}{q_2}}}{{{r^2}}}$

where ${\varepsilon _0}$ = absolute permittivity of free space.

The charge on a proton is equal to –

$e = 1. 602 \times {10^{ - 19}}C$

The weight of the proton is equal to the product of mass and acceleration due to gravity.

$W = mg$

Mass of electron, $m = 1. 67 \times {10^{ - 27}}kg$

Acceleration due to gravity, $g = 10m{s^{ - 2}}$

Weight, $W = 1. 67 \times {10^{ - 27}} \times 10 = 1. 67 \times {10^{ - 26}}N$

Given that the force acting between the protons is equal to the weight of a proton, we have

$F = W$

$ \Rightarrow \dfrac{1}{{4\pi {\varepsilon _0}}}\dfrac{{{q_1}{q_2}}}{{{r^2}}} = W$

Substituting, we have –

$9 \times {10^9} \times \dfrac{{{{\left( {1. 6 \times {{10}^{ - 19}}} \right)}^2}}}{{{r^2}}} = 1. 67 \times {10^{ - 26}}$

$ \Rightarrow {r^2} = \dfrac{{{{\left( {1. 6 \times {{10}^{ - 19}}} \right)}^2} \times 9 \times {{10}^9}}}{{1. 67 \times {{10}^{ - 26}}}}$

$ \Rightarrow {r^2} = \dfrac{{2. 56 \times {{10}^{ - 38}} \times 9 \times {{10}^9}}}{{1. 67 \times {{10}^{ - 26}}}}$

\[ \Rightarrow {r^2} = \dfrac{{2. 56 \times 9}}{{1. 67}} \times {10^{ - 38 + 9 + 26}}\]

\[ \Rightarrow {r^2} = 13. 796 \times {10^{ - 3}} = 0. 0137\]

\[\therefore r = \sqrt {0. 0137} = 0. 117m\]

**Hence, the distance of separation is equal to 0.117m.**

**Note:**The proton and electron have the same charge, known as the fundamental charge $e = 1. 602 \times {10^{ - 19}}C$, but they are opposite in sign. While proton has charge equal to $ + e$, the charge of electrons is $ - e$. However, in this question, only the magnitude of the charge is considered in the calculations and not the sign.

Recently Updated Pages

Basicity of sulphurous acid and sulphuric acid are

Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred

What is the stopping potential when the metal with class 12 physics JEE_Main

The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main

How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE

Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE

Trending doubts

How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Describe the poetic devices used in the poem Aunt Jennifers class 12 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Difference Between Plant Cell and Animal Cell

Change the following sentences into negative and interrogative class 10 english CBSE

State the laws of reflection of light

State and prove Bernoullis theorem class 11 physics CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE