In an experiment done by scattering 5.5 MeV α particleson a thin gold foil, students find that 10,000 α particles arescattered at an angle greater than 90 degrees.
How many α particles will be scattered between 60 degreesand 80 degrees?
To calculate the number of α particles scattered between 60 degrees and 80 degrees, we need to use the Rutherford scattering formula.
Rutherford scattering formula:
dσ/dΩ = (Z^2 e^4 / (16πε_0)^2) * (1 / (4E^2 sin^4(θ/2))) * (1 / sin^4(θ/2))
where,
dσ/dΩ = differential scattering cross-section
Z = atomic number of gold (79 for Au)
e = elementary charge (1.6 x 10^-19 C)
ε_0 = permittivity of free space (8.85 x 10^-12 C^2 / Nm^2)
E = kinetic energy of the α particles (5.5 MeV = 8.8 x 10^-13 J)
θ = scattering angle in radians
First, let's convert the energy of α particles into joules:
5.5 MeV = 5.5 x 10^6 eV = 5.5 x 10^6 x 1.6 x 10^-19 J
= 8.8 x 10^-13 J
Now, we'll calculate the number of α particles scattered between 60° and 80° using the equation:
N = n x σ x A
Where,
N = Number of α particles scattered
n = Number of incident α particles (initially 10,000)
σ = Differential scattering cross-section
A = Solid angle subtended by the range of scattering angles (in steradians)
To calculate the differential scattering cross-section, we'll use the Rutherford scattering formula and integrate it over the range of angles (θ):
σ = ∫(dσ/dΩ) dΩ
Since the differential cross-section is a function of θ, we can rewrite this integral as:
σ = ∫[dσ/dθ * (dθ/dΩ)] dΩ
The solid angle, A, subtended by the range of scattering angles (60° to 80°) is given by:
A = 2π * (cos(60) - cos(80)) = 2π * (0.5 - 0.1736)
Now, let's substitute the values into the equations and calculate:
σ = ∫[(Z^2 e^4 / (16πε_0)^2) * (1 / (4E^2 sin^4(θ/2))) * (1 / sin^4(θ/2))] dΩ
A = 2π * (cos(60) - cos(80))
Finally, we'll multiply the obtained value of σ with n (initially, 10,000) to get the number of α particles scattered between 60° and 80°:
N = n x σ x A
To determine the number of α particles scattered between 60 degrees and 80 degrees, we need to use the Rutherford scattering formula. This formula relates the number of scattered particles to the angle of deflection and the scattering cross-section. Here's how you can calculate it:
Step 1: Start with the Rutherford scattering formula:
dN = (N * n * Z^2 * e^4) / (16 * ε0^2 * E^2 * sin^4(θ/2)) * dΩ
Where:
dN is the number of particles scattered into a solid angle dΩ.
N is the total number of incident particles.
n is the number density of scattering centers (atoms/cm^3).
Z is the atomic number of the target material (gold has Z = 79).
e is the elementary charge (1.602 x 10^(-19) C).
ε0 is the vacuum permittivity (8.854 x 10^(-12) C^2/(N.m^2)).
E is the kinetic energy of the incident particles (5.5 MeV).
θ is the scattering angle.
Step 2: We need to find the ratio of scattered particles at angles between 60 and 80 degrees to the total scattered particles at angles greater than 90 degrees.
ratio = (dN_60_80 / dN_90plus)
Step 3: Plug in the values into the formula. Since we need to find the ratio, we can cancel out terms that are common in both equations. The terms that will cancel out are: N, n, Z, e^4, ε0^2, and E^2.
ratio = sin^(-4)(θ_60) / sin^(-4)(θ_90)
Step 4: Substitute the angles into the formula.
ratio = sin^(-4)(60) / sin^(-4)(90)
Step 5: Calculate the ratio.
ratio = (sin(60))^(-4) / (sin(90))^(-4)
ratio = (0.866)^(-4) / (1)^(-4)
ratio = 11.52
Step 6: Finally, multiply the ratio by the total number of particles scattered at angles greater than 90 degrees (10,000 α particles).
particles_60_80 = ratio * particles_90plus
particles_60_80 = 11.52 * 10,000
particles_60_80 ≈ 115,200 α particles
Therefore, approximately 115,200 α particles will be scattered between 60 degrees and 80 degrees in this experiment.