Sparks occur when the electric field in air exceeds 3 x 106 N/C. This is because free electrons normally present in air are accelerated to such high speeds that their kinetic energy will overcome the potential energy holding other electrons to atoms. When those electrons rearrange themselves after such a collision, a flash of light is emitted. Let us suppose that the work done on an electron must give it an energy of 3 x 10-19 J to cause this ionization. How far does an electron involved in making in a spark travel through the air before it collides with an atom

Respuesta :

Answer:

h = 5.38 10¹⁶ m

Explanation:

Let's start this exercise by assuming that all the potential energy of the electron is converted into kinetic energy, let's use the conservation of energy

starting point. Just before ionization

          Em₀ = U = qE

final point. Right after ionization

           Em_f = K = ½ m v²

Energy is conserved

           Em₀ = Em_f

           q E = ½ m v²

           v² = 2qE / m

Now we can use the relationship between net work and kinetic energy

           W_net = ΔK

net work is the work done by the electron minus the binding energy with the atom, called the work function, Ф = 3 10-19 J

           W - Ф = K_f - K₀

we assume that the electron converts all its initial initial kinetic energy to be zero

           W -Ф = ½ m v² - 0

            W = ½ m v² +Ф

we substitute

           W = 1/2 m 2qE/m + E

           W =  qE +Ф

           W = 1.6 10⁻¹⁹ 3 10⁶ + 3 10⁻¹⁹

            W = 4.8 10⁻¹³ + 3 10⁻¹⁹

           W = 4.8 10⁻¹³ J

When the electron is in air, its kinetic energy can be transformed into gravitational potential energy

           

As the electron is in the air, all work is transformed into scientific energy

           W = K

starting point Em₀ = K = W

end point Em_F = U = m g h

energy conservation Em₀ = Em_f

                       W = m g h

                       h = [tex]\frac{W}{mg}[/tex]

let's calculate

                       h = [tex]\frac{4.8 \ 10^{-13} x}{9.1 \ 10^{-31} \ 9.8 }[/tex]

                       h = 5.38 10¹⁶ m

Electron involved in making in spark travel through the air before it collides with an atom will be at the distance of 5.38 10¹⁶ m.

What is an electric field?

An electric field is an electric property that is connected with any location in space where a charge exists in any form. The electric force per unit charge is another term for an electric field.

Let's begin this exercise by assuming that all of the electron's potential energy is turned into kinetic energy, and then we'll apply the law of conservation of energy.

Energy before ionization;

[tex]\rm Em_0 = U = qE[/tex]

Energy after ionization;

[tex]Em_f = K = \frac{1}{2} mv^2[/tex]

From the law of conservation of energy principle;

[tex]Em_0 = Em_f \\\\ q E =\frac{1}{2} m v^2\\\\ v^2 = \frac{2qE }{m}[/tex]

The relationship between net work and kinetic energy;

[tex]W_{net} = \triangle K[/tex]

The work function is defined as net work, which is the work done by the electron minus the binding energy with the atom.

[tex]W - \phi = K_f - K_0[/tex]

[tex]W = K_f+ \phi[/tex]

[tex]W = \frac{1}{2} m \times \frac{2qE}{m} + E\\ \\W = qE + \phi \\\\ \rm W = 1.6 \times 10^{-19}\times 3 \tims 10^6 3 10⁶ +3 \times 10^{-19} \\\\ W = 4.8 \times 10^{-13}+ 3 \times 10^{-19}\\\\ W = 4.8 \times 10^{-13} J[/tex]

EMF at starting point;

[tex]\rm Em_0 = K = W[/tex]

EMF at the endpoint;

[tex]\rm Em_F = U = m g h[/tex]

From the law of conservation of energy principle;

[tex]Em_0 = Em_f \\\\ W = m g \\\\ h = \frac{W}{mg}\\\\\ h = \frac{4.8 \timjes 10^{-13}}{9.1 \times 10^{-31} \times 9.81 }\\\\ \rm h= 5.38 \times 10^{16}[/tex]

Hence electron involved in making in spark travel through the air before it collides with an atom will be at a distance of 5.38 10¹⁶ m.

To learn more about the electric field refer to the link;

https://brainly.com/question/26690770