What fraction of 5 MeV α particles will be scattered through angles greater than 8.5° from a gold foil (Z = 79, density = 19.3 g/cm3) of thickness 10-8 m?

And after do all the operations we got:

And that would be the final answer for this case.

Explanation:

For this case we can use the fomrula for the fraction of incident particles scattered by an angle , given by:

Where:

represent the charge of the projectile (Z1=2)

is the target charge (z2=79)

represent the kinetic energy of incident particle

n represent the density of target particles (we need to find it first)

represent the thicknss of the foil

The first step would be calculate the density of target particles with the following formula:

Where:

the Avogadro's number

represent the atoms per molecule

represent the molecular weigth

If we replace we got:

Now we can calculate the fraction of 5 MeV alpha particles that would be scatteres with angle higher than 8 degrees in a piece of thickness

And using the first formula we got:

And after do all the operations we got:

And that would be the final answer for this case.

v=ir = 3x15 =45 v