Electron scattering find acceleration

[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 doesn't work out. I am not sure if I am using the correct values for the q and Q though.

 

[Astronuc]

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

 

[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 can't figure out what it would be.

 

[Astronuc]

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

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

 

[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?

 

[Astronuc]

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?

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.