Answer:
Vf = 1.43 m/s
Step-by-step explanation:
From Coulomb's Law, the electrostatic force between electron and proton is given as:
F = kq₁q₂/r²
F = Electrostatic force = ?
k = Coulomb's Constant = 9 x 10⁹ N.m²/C²
q₁ = magnitude of charge on electron = 1.6 x 10⁻¹⁹ C
q₂ = magnitude of charge on proton = 1.6 x 10⁻¹⁹ C
r = distance between electron and proton = 9 cm = 0.09 m
Therefore,
F = (9 x 10⁹ N.m²/C²)(1.6 x 10⁻¹⁹ C)(1.6 x 10⁻¹⁹ C)/(0.09 m)²
F = 2.84 x 10⁻²⁶ N
but, from Newton's second law:
F = 2.84 x 10⁻²⁶ N = ma
where,
m = mass of electron = 9.1 x 10⁻³¹ kg
a = acceleration of electron = ?
Therefore,
2.84 x 10⁻²⁶ N = (1.67 x 10⁻²⁷ kg)(a)
a = 2.84 x 10⁻²⁶ N/1.67 x 10⁻²⁷ kg
a = 17.03 m/s²
Now, we apply 3rd equation of motion to the motion of electron from a distance of 9 cm to 3 cm near to the proton:
2as = Vf² - Vi²
where,
s = distance traveled = 9 cm - 3 cm = 6 cm = 0.06 m
Vf = speed of electron when it is 3 cm from proton = ?
Vi = Initial speed of electron = 0 m/s
Therefore,
2(17.03 m/s²)(0.06 m) = Vf² - (0 m/s)²
Vf = √2.04 m²/s²
Vf = 1.43 m/s