Respuesta :
Answer:
[tex] 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) [/tex]
And after do all the operations we got:
[tex] f(8) = 1.96 x10^{-4] atoms/m^2[/tex]
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 [tex]\theta[/tex], given by:
[tex] f(\theta) = \pi nt (\frac{Z_1 e Z_2 e}{8 \pi e_o K})^2 cos^2 (\theta/2)[/tex]
Where:
[tex] Z_1 e [/tex] represent the charge of the projectile (Z1=2)
[tex] Z_2 e[/tex] is the target charge (z2=79)
[tex]K= 5x10^6 eV[/tex] represent the kinetic energy of incident particle
n represent the density of target particles (we need to find it first)
[tex]t= 10^{-8] m[/tex] represent the thicknss of the foil
The first step would be calculate the density of target particles with the following formula:
[tex] n =\frac{\rho N_A N_M}{M_g}[/tex]
Where:
[tex] \rho = 19.3 g/m^3= 19300 Kg/m^3[/tex]
[tex] N_A = 6.022 x10^{23} molecules/mol[/tex] the Avogadro's number
[tex] N_M = 1[/tex] represent the atoms per molecule
[tex] M_g = 197 g/mol = 0.197 Kg/mol[/tex] represent the molecular weigth
If we replace we got:
[tex] 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[/tex]
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 [tex] t=10^{-8}m[/tex]
And using the first formula we got:
[tex] 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) [/tex]
And after do all the operations we got:
[tex] f(8) = 1.96 x10^{-4] atoms/m^2[/tex]
And that would be the final answer for this case.
