Distance of closest approach of alpha particle calc

[Physgeek64]

Homework Statement

An alpha particle is accelerated from rest through a potential difference of 20kV. It travels directly towards
a stationary Beryllium nucleus (4 protons, 5 neutrons). Calculate the distance of closest approach

Homework Equations

The Attempt at a Solution

E=qv=(40x10^3)e, where e is the electronic charge
U=qQ/4(pi)(epsilon naught)(R)
R=qQ/4(pi)(epsilon naught)(U)
R=2.876x10-13 m

Is there any more to this question- it seems a bit to simple for it to be just this. Wondering if I've missed something blindingly obvious. Thanks :)

 

[Fightfish]

What is the speed of the particle at the point of closest approach?

 

[Physgeek64]

What is the speed of the particle at the point of closest approach?

0- so i assume all of its k.e is converted into potential energy?

 

[SammyS]

0- so i assume all of its k.e is converted into potential energy?

That's not so much an assumption as it is the result of applying conservation of energy.

... and, yes, that's correct.

 

[gleem]

Is there any more to this question- it seems a bit to simple for it to be just this. Wondering if I've missed something blindingly obvious. Thanks :)

You made an inadvertent assumption about this scattering processes.

The mass of the alpha particle is of the same order of magnitude as that of the Beryllium atom. So what happens to the Beryllium atom as the alpha particle approaches it?

 

[Physgeek64]

You made an inadvertent assumption about this scattering processes.

The mass of the alpha particle is of the same order of magnitude as that of the Beryllium atom. So what happens to the Beryllium atom as the alpha particle approaches it?

Well I would have thought it would be itself 'pushed' backwards. But I don't really see how to work that into my equations. Especially since this will cause the force on the alpha particle to not follow the inverse square law.

Ive tried applying Newtons laws, but it just gets very messy since the force is changing

Doing this I get

q1*q2/4*pi*e0*(x+r)^2=m(d^2x/dt^2)

Then applying the conservation of momentum I get

m(a)*v(0)=m(a)(dr/dt)+m(b)(dx/dt)

And v(0) can be worked out using the conservation of energy as E=qV=1/2mv(0)^2

So this expression for momentum can be integrated to get r in terms of x to get

r=v(0)*t-(m(b)/m(a))*x

Then this can be subbed back into the equation of motion for x to get

K*(x*(1-m(b)/m(a))+v(0)t)^-2=m*(d^2x/dt^2)

Where K=q1*q2/4*pi*e0

But I'm not really sure how to go about solving this since its second order and both t and x are present

 

[gneill]

A handy trick is to convert the frame of reference from the stationary "lab" frame to the center of momentum frame of reference. It's a coordinate system that moves with the center of mass. In such a frame the point where the two objects would collide (if it weren't prevented by their mutual repulsion) is the center of mass and it is stationary. In this frame the two approaching particles are also stationary at the instant of closest approach.

 

[Physgeek64]

A handy trick is to convert the frame of reference from the stationary "lab" frame to the center of momentum frame of reference. It's a coordinate system that moves with the center of mass. In such a frame the point where the two objects would collide (if it weren't prevented by their mutual repulsion) is the center of mass and it is stationary. In this frame the two approaching particles are also stationary at the instant of closest approach.

Ah okay! Thats really helpful! Thank you- I shall give that a go :)