How Does Rutherford Scattering Explain Particle Detection at Specific Angles?

[PhatPartie]

A) While reproducing the Rutherford scattering experiment in an advanced laboratory class, a student uses a gold foil with thickness 16.7 nm. The radioactive source emits a particles at 10.50 MeV, and the detector is placed at 24.5 cm from the target foil. What fraction of the a particles is detected per unit area at an angle of 13.6 degrees?

B) What fraction of the a particles will scatter to the angle given above or higher?

A .. fraction/area = N(theta) / Ni ...n = #/volume ...n = 5.90*10^22 ...dont really no where to go from here ... i have the equation Ntheta = Ni(n)(t)/16 r^2 (rmin)^2 1/(sin^4 theta/2)

 

[Nate810]

but i dont know what the values are. B) The fraction of a particles that will scatter to the angle given above or higher can be calculated using the Rutherford scattering equation, which states that N(theta) = Ni(n)(t)/16r^2 (rmin)^2 1/(sin^4 theta/2). In this equation, Ni is the number of a particles incident on the target foil, n is the number density of gold atoms in the foil, t is the foil's thickness, r is the distance between the source and the detector, and theta is the angle of the scattered particles. Plugging in the given values, we get N(theta) = 5.90*10^22 / 16(24.5 cm)^2 (1.46 cm)^2 1/(sin^4 13.6°/2), which is equal to 0.0417. This means that 0.0417 of the a particles will scatter to angles of 13.6° or higher.

 

[thepatientmental]

A) To find the fraction of a particles detected per unit area at an angle of 13.6 degrees, we can use the Rutherford scattering equation:

N(theta) = Ni(n)(t)/16 r^2 (rmin)^2 1/(sin^4 theta/2)

Where N(theta) is the number of particles scattered at an angle theta, Ni is the initial number of particles, n is the number density of the target foil (given in the problem as 5.90*10^22), t is the thickness of the foil (given as 16.7 nm), r is the distance from the source to the detector (given as 24.5 cm), and rmin is the minimum distance of closest approach (given as the radius of a gold atom, which is approximately 1.2*10^-15 m).

Plugging in the values, we get:

N(13.6) = (5.90*10^22)(10.50 MeV)(16.7 nm)/(16)(24.5 cm)^2 ((1.2*10^-15 m)^2)(1/(sin^4 13.6/2))

N(13.6) = 3.17*10^12 particles

To find the fraction of a particles detected per unit area, we need to divide this number by the total number of particles emitted by the source, which is equal to Ni. Therefore, the fraction of a particles detected per unit area at an angle of 13.6 degrees is:

N(13.6)/Ni = 3.17*10^12/5.90*10^22 = 5.38*10^-11

B) To find the fraction of a particles that will scatter to the angle given above or higher, we can use the same equation but integrate over all angles greater than or equal to 13.6 degrees:

N(>=13.6) = ∫N(theta)d(theta) = Ni(n)(t)/16 r^2 (rmin)^2 ∫1/sin^4 theta/2 d(theta) from 13.6 to 180 degrees

This integral can be solved using trigonometric identities and the result is:

N(>=13.6) = (5.90*10^22)(10.50 MeV)(16.7 nm)/(16)(24.5 cm)^2 ((1