# Electron scattering find acceleration

1. Oct 9, 2006

### fishys182

Lead nucleus has charge = +82e
and raadius R = 7.10*10^(-15) m
permittivity of free space = 8.85*10^(-12) C^2/Nm^2
magnitude of charge on electron e = 1.60*10^(-19) C
mass of electron = 9.11*10^(-31) kg

find the acceleration 4R from the center of the lead nucleus.

how do i do this?

F = kqQ/(r^2) then a = F/m doesnt work out. im not sure if im using the correct values for the q and Q though.

2. Oct 9, 2006

### Staff: Mentor

a = F/m should work.

What values are you using for Q (nucleus) and q (electron)?

For electron q = e = -1.60*10^(-19) C

For the nucleus, it depends if one is considering the shielding of the electrons or not. Q on the nucleus = Ze = +82e = 82*1.60*10^(-19) C.

At 4R, where R is the effective radius of the nucleus, the electron probably experiences the full coulombic field of the nucleus.

And in the denominator, r2 would be (4R)2

3. Oct 9, 2006

### fishys182

thank you.

i got to this point, but now i am confused about how it would be different if it were R/4 instead of 4R.

the equation is different i think, but i cant figure out what it would be.

4. Oct 9, 2006

### Staff: Mentor

If r = 4R, then r2 = 16R2, and 1/r2 = 1/(16R2)

If r = R/4, then r2 = R2/16, and 1/r2 = 16/R2.

5. Oct 9, 2006

### fishys182

hmm
i entered that in before though, and it said my answer was incorrect.

isnt there a difference for when r>=R and r<R?

6. Oct 9, 2006

### Staff: Mentor

Well, yes. When r < R, then the electron is within the nucleus and it would be interacting with a completely different and more complex charge field than outside the nucleus.

http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/elescat.html

http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/scatele.html

Have you solved the classical EM problem for an electric field in a sphere of uniformly distributed charge and compared to the E-field outside? The nucleus is more complex.

Last edited: Oct 9, 2006