Question

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?

Answer 1

Answer:

$f(8) = \pi (5.90x10^{28} atoms/m^3) *(10^{-8} m) (\frac{2*79*(1.6x10^{-19}C)^2}{8 \pi (8.85 x10^{-12} C^2/Nm^2)*(5x10^6 Nm) *1.6x10^{-19}C})* cot^2 (8/2)$

And after do all the operations we got:

$f(8) = 1.96 x10^{-4] atoms/m^2$

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 $\theta$, given by:

$f(\theta) = \pi nt (\frac{Z_1 e Z_2 e}{8 \pi e_o K})^2 cos^2 (\theta/2)$

Where:

$Z_1 e$ represent the charge of the projectile (Z1=2)

$Z_2 e$ is the target charge (z2=79)

$K= 5x10^6 eV$ represent the kinetic energy of incident particle

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

$t= 10^{-8] m$ represent the thicknss of the foil

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

$n =\frac{\rho N_A N_M}{M_g}$

Where:

$\rho = 19.3 g/m^3= 19300 Kg/m^3$

$N_A = 6.022 x10^{23} molecules/mol$ the Avogadro's number

$N_M = 1$ represent the atoms per molecule

$M_g = 197 g/mol = 0.197 Kg/mol$ represent the molecular weigth

If we replace we got:

$n = \frac{19300 kg/m^3 *6.022x10^{23} molecu/mol * 1 atom/mole}{0.197 Kg/mol}= 5.90 x106{28] atoms/m^3$

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 $t=10^{-8}m$

And using the first formula we got:

$f(8) = \pi (5.90x10^{28} atoms/m^3) *(10^{-8} m) (\frac{2*79*(1.6x10^{-19}C)^2}{8 \pi (8.85 x10^{-12} C^2/Nm^2)*(5x10^6 Nm) *1.6x10^{-19}C})* cot^2 (8/2)$

And after do all the operations we got:

$f(8) = 1.96 x10^{-4] atoms/m^2$

And that would be the final answer for this case.