An electron (mass 9 * 10-31 kg) is traveling at a speed of 0.92 in an electron accelerator. An electric force of 1.4 * 10-13 N is applied in the direction of motion while the electron travels a distance of 3 m. You need to find the new speed of the electron.

Given Information:

Mass of electron = m = 9x10⁻³¹ kg

initial speed of electron = v₁ = 0.92c

Force = F = 1.4x10⁻¹³ J

Distance = d = 3 m

Required Information:

Final speed of electron = v₂ = ?

Answer:

Final speed of electron = v₂ = 2.974x10⁸ m/s

Explanation:

As we know from the conservation of energy,

E₂ - E₁ = W

E₂ = E₁ + W

Where E₂ is the final energy of electron and E₁ is the initial energy of electron

The above equation can be written in the form of particle energy

γ₂mc² = γ₁mc² + W

where γ₁ and γ₂ are given by

γ₁ = 1/√1 - (v₁/c) ²

γ₂ = 1/√1 - (v₂/c) ²

First calculate γ₁

γ₁ = 1/√1 - (0.92c/c) ²

γ₁ = 2.55 m

Now calculate γ₂

γ₂ = (γ₁mc² + W) / mc²

First we need to find the work done

W = F*d

W = 1.4x10⁻¹³*3

W = 4.2x10⁻¹³ J

so γ₂ is

γ₂ = (2.55*9x10⁻³¹ * (3x10⁸) ² + 4.2x10⁻¹³) / 9x10⁻³¹ * (3x10⁸) ²

γ₂ = 7.73

Now we can find the new speed of the electron

γ₂ = 1/√1 - (v₂/c) ²

Re-arranging the above equation results in

v₂ = c*√ (1 - 1/γ₂²)

v₂ = 3x10⁸*√ (1 - 1/7.73²)

v₂ = 2.974x10⁸ m/s