Question

What fraction of 5 MeV α particles will be scattered through angles greater than 2.5° from a gold foil (Z = 79, density = 19.3 g/cm3) of thickness 10-8 m?

Answer 1

We know that no. of atoms per cubic meter
n = pN_a/M
Where N_a = 6.022*10²³ is avogadro's no.
p is density of gold = 19.3 g/cm³ = 19.3*10⁶ g/m³
M is molecular mass of gold = 196.97 g/mole
n = [(19.3*10⁶ )*( 6.022*10²³)/(196.97)] = 5.9*10²⁸ atoms/m³
Now fraction of particles scattered by angle

Where E is energy of particle = 5*10⁶*(1.6*10^-19) = 8*10^-13 J
z₁ is atomic no. of gole = 79
Z₂ is atomic no. of particle = 2
t is thickness of sheet = 10^-8 m

f = (1.85*10²¹)*(5.167*10^-28)*(2100.329)
f = 2.008*10^-3