so are u saying u cud use
F = KQ1Q1/r2 which will give r2 = KQ1Q2/F ???? ( providing the force is found )
exactly.Find the force by measuring the angle.
lol in an exam they will never ever ever say " how do u find the closest approach of an alpha particle to a nucleus? " wat a load of ****.
if b is the impact parameter (i.e. the displacement of the alpha particle from the parallel head-on axis when it is fired), Z is the charge of the nucleus, m the mass of the alpha particle, u its initial speed, v its speed at closest approach, R the distance of closest approach, e the magnitude of the charge of an electron, and k the electrostatic constant ( = 1/4πε ):
conservation of ang. momentum: mub = mvR
conservation of energy: mu²/2 = mv²/2 - 2kZe/R²
you have 2 equations in 2 unknowns, solve for R.
where V is the electric potential in volts (or JOULES per COULOMB OF CHARGE) Q is the charge of the nucleus and r is the distance from the centre of the nucleus ε0 and π are your constants.
The potential energy of an alpha particle at that point is, therefore, given by:
E= 1/4πε0 * Qq/r
where q is the charge on the alpha particle.
Therefore to get an alpha particle to that point you need to put E Joules in.
Conversely if you give an alpha E joules of kinetic energy (they are all emitted with the same Kinetic Energy I am told) the distance of closest approach is given by:
d= (Qq/E*4πε0)-radius of nucleus
(I think- my algebra is sketchy at best- especially on a computer- seems easier on paper).
IF you assume that the particle touches the nucleus then you can use this distance (or rather the distance given above but without the "-radius of nucleus" bit) to estimate the size of the nucleus. It certainly provides an upper limit.
Hope this helps (hope I'm right too).
(by the way you know the Atomic number of the nucleus because you know what you are shooting the alpha particle at e.g a bit of gold foil gives an atomic number 79 (79*1.6e-19C of charge).