An alpha particle is a nucleus of helium.It has twice the charge and four times the mass of the proton.When they were very far away from each other,but headed toward directly each other,a proton and an alpha particle each had an initial speed of 0.0030c,where c is the speed of light.What is their distance of closest approach? Hint: There are two conserved quantities.Make use of both of them.(c = 3.00 × 10⁸ m/s,k = 1/4πε₀ = , , A) 2.1 × 10^-13 m B) 3.3 × 10^-13 m C) 2.6 × 10^-13 m D) 2.9 × 10^-13 m

Question

Quizplus · Accepted Answer

We can use conservation of energy and conservation of angular momentum to solve this problem. The total energy of the system is conserved, and at the point of closest approach, all of the kinetic energy will have been converted into potential energy. Let E be the total energy of the system:

E = K + U

where K is the kinetic energy and U is the potential energy. At the point of closest approach, K = 0, so we have:

E = U

The potential energy can be calculated using Coulomb's law:

U = -k(Q1Q2/r)

where k is the Coulomb constant, Q1 is the charge of particle 1, Q2 is the charge of particle 2, and r is their separation. For the given problem, we have Q1 = +e (charge of proton) and Q2 = +2e (charge of alpha particle), where e is the elementary charge. Thus:

U = -k(e)(2e)/r = -2ke^2/r

The angular momentum of the system is also conserved. Since the particles are moving directly toward each other, the only component of angular momentum that is relevant is the z-component. We can write:

Lz = r x p

where r is the position vector relative to the center of mass of the system and p is the momentum. Since the system is moving directly toward each other, we can assume that the z-component of r is initially zero. The initial z-component of p is:

pz = mv = (mp+v)a

where mp is the mass of the proton, v is its initial speed, a is the acceleration due to the attraction from the alpha particle, and (mp+v) is the total mass of the proton and its recoil due to the attraction. We can write:

a = kQ2/mpd^2 = 2ke^2/(mpd^2)

where d is the initial separation between the particles. Thus:

pz = (mp+v)(2ke^2/mpd^2) = 2ke^2/d

The magnitude of the angular momentum is Lz = rpz, so we have:

Lz = r(2ke^2/d)

Using conservation of angular momentum, we have:

Lz,i = Lz,f

mrivi = rpz

where m is the reduced mass of the system (mp/3), vi is the initial speed of the system (both particles have the same initial speed), and r is the distance of closest approach. Thus:

r = mpvi/3pz

Substituting the expressions for pz and U, we have:

r = mpvi/3(2ke^2/d)(-2ke^2/r)

Solving for r, we get:

r = 2ke^2/mpvi^2

Substituting the given values, we get:

r = (2)(9x10^9 Nm^2/C^2)(1.6x10^-19 C)^2/(1.67x10^-27 kg)(0.0030c)^2

r = 2.1x10^-13 m

Therefore, the answer is A.

An Alpha Particle Is a Nucleus of Helium