Answer: Vf = 2,400,000 m/s
Explanation:
1) The only relevant force is the electrostatic force
2) The formula for the electrostatic force is F = E×q
Where E is the electric field and q is the magnitude of the charge.
3) Since the electric field is the same in both cases, and the charge of the protons and electrons have the same magnitude, you can state that the magntitude of the electric forces acting in both proton and electron are the same
4) Fe = Fp (Fe stands for force on the electron and Fp stands for force on the proton).
5) Using second law of Newton, Force = mass × acceleration
Fe = Me × Ae (Me is mass of the electron, and Ae is acceleration of the electron)
Fp = Mp × Ap (Mp is mass of the proton, and Ap is acceleration of the proton)
⇒Me × Ae = Mp × Ap
⇒ Ae = Mp × Ap / Me
6) Now, state the equations for the velocity in uniformly accelerated motion:
i) Vf² = Vo² + 2ad
Vo² = 0 for both cases, and d is the same distance.
⇒ Vf² = 2ad
ii) For the proton Vf² = 2(Ap)(d) ⇒ Ap = Vf² / (2d)
⇒ Ap = (55,000 m/s)² / (2d)
iii) For the electron Vf² = 2(Ae)² (2d)
iv) Using Ae = Mp × Ap / Me (found prevously):
Vf² = Mp × (55,000 m/s)² / (2d) × (2d) / Me
⇒ Vf² = Mp × (55,000 m/s)² / Me
Taking square root in both sides:
⇒ Vf = 55,000 m/s × √ [Mp / Me]
7) These are the values for the masses of a proton and an electron:
Mp = 1.67 × 10⁻²⁷ kg
Me = 9.11×10⁻³¹ kg
8) Replace and compute:
Vf = 55,000 m/s × √ [ 1.67 × 10⁻²⁷ kg / 9.11×10⁻³¹ kg] = 2,354,841.8 m/s
Round to two significan digits: Vf = 2,400,000 m/s
