Question

A proton and an alpha particle are released from rest when they are 0.225 nm apart. The alpha particle (a helium nucleus) has essentially four times the mass and twotimes the charge of a proton.

a) Find the maximum speed of proton.

b) Find the maximum speed of alpha particle.

c) Find the maximum acceleration of proton.

d) Find the maximum acceleration of alpha particle.

Answer 1

Equation needed:

Ki + Ui = Kf + Uf

Kinetic energy = (1/2)mv²

^{Qa = charge of alpha particle = 2*Qp}

^{Qp = charge of proton particle = 1.60E-19 C}

When looking at equations regarding electrical charges, Ki + Ui = Kf + Uf is the only one that has velocity involved - Kinetic energy = (1/2)mv²

Momentum is reserved: 0 = mpvp - mava , mpvp = mava → va = (mpvp)/ma

a) Find max speed of proton:

Ki = 0 because the particle was released from rest

Uf = 0 because all the potential energy will be used up

thus, Ui = Kf . U = Vq = (kq/r)q = kQpQa/r . This goes the same for alpha.

Ui = K(proton) + K(alpha)

[k(2Qa)(Qa)]/r = (1/2)mpvp² + (1/2)mava²

[(9E9)(2)(1.60E-19)² ]/0.225E-9 = (1/2)(1.67E-27)(vp)² + (1/2)(4*1.67E-27)(mpvp/ma)² 

2.048E-18 = (8.35E-28)vp² + (2.088E-28)vp²

2.048E-18 = (1.044E-27)vp²

velocity of proton particle (vp) = 4.43E4 m/s

 

b. Find max speed of alpha particle:

According to va = (mpvp)/ma , va = 1.107E4 m/s

 

c. Find accerlation of proton particle:

F = ma → a = F/m .......................... F = Eq → F = (kQaQp)/(r²)

F = [(9E9)(2)(1.60E-19)² ]/(0.225E-9)²

F = 9.102E-9 N

Accerlation of proton (a) = 9.102E-9/1.67E-27 = 5.45E18 m/s²

 

d. Find acceleration of alpha particle:

• Same method as (c) but with mass of alpha.

Acceleration of alpha (a) = 1.36E18 m/s²