Answer:
K.E = 15.57 x 10⁻¹⁷ J
Explanation:
First, we find the acceleration of the electron by using the formula of electric field:
E = F/q
F = Eq
but, from Newton's 2nd Law:
F = ma
Comparing both equations, we get:
ma = Eq
a = Eq/m
where,
E = electric field intensity = 120 N/C
q = charge of electron = 1.6 x 10⁻¹⁹ C
m = Mass of electron = 9.1 x 10⁻³¹ kg
Therefore,
a = (120 N/C)(1.6 x 10⁻¹⁹ C)/(9.1 x 10⁻³¹ kg)
a = 2.11 x 10¹³ m/s²
Now, we need to find the final velocity of the electron. Using 3rd equation of motion:
2as = Vf² - Vi²
where,
Vf = Final Velocity = ?
Vi = Initial Velocity = 1.4 x 10⁷ m/s
s = distance = 3.5 m
Therefore,
(2)(2.11 x 10¹³ m/s²)(3.5 m) = Vf² - (1.4 x 10⁷)²
Vf = √(1.477 x 10¹⁴ m²/s² + 1.96 x 10¹⁴ m²/s²)
Vf = 1.85 x 10⁷ m/s
Now, we find the kinetic energy of electron at the end of the motion:
K.E = (0.5)(m)(Vf)²
K.E = (0.5)(9.1 x 10⁻³¹ kg)(1.85 x 10⁷ m/s)²
K.E = 15.57 x 10⁻¹⁷ J
