<h2>
Answer:</h2>
1.45 x 10⁷m/s
<h2>
Explanation:</h2>
In any system, the conservation of mechanical energy is always conserved. In other words, the sum of potential () and kinetic energy () is constant. i.e
+ = k
=> Δ + Δ = 0
=> Δ = -Δ ---------------------(i)
The system considered here is an electric field.
(i) We know that the change in potential energy, Δ, of a charge Q as it moves between two points in an electric field is given as;
Δ = Q x Δ V ---------------(ii)
Where;
ΔV = change in electric potential as the charge moves negative to positive plate = 600V
Q = charge of an electron = -1.6 x 10⁻¹⁹C
Substitute these values into equation (ii)
Δ = -1.6 x 10⁻¹⁹C x 600V = -9.6 x 10⁻¹⁷J
(ii) Also, since the electron is starting from rest, the change in kinetic energy is given by;
= x m x v² -------------------(ii)
Where;
m = mass of the electron = 9.1 x 10⁻³¹ kg
v = final velocity of the electron
Substitute these values into equation (ii) as follows;
= x 9.1 x 10⁻³¹ x v² J
(iii) Now substitute the values of = x 9.1 x 10⁻³¹ x v² J and Δ = -9.6 x 10⁻¹⁷J into equation (i) as follows;
-9.6 x 10⁻¹⁷ = - x 9.1 x 10⁻³¹ x v²
2 x 9.6 x 10⁻¹⁷ = 9.1 x 10⁻³¹ x v²
v² =
v² = 2.11 x 10¹⁴
v =
v = 1.45 x 10⁷ m/s
The electron is going as fast as 1.45 x 10⁷m/s