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An iron foil of thickness 17μm is used in a Rutherford experiment to scatter α particles with energy 11MeV. (a) Determine the impact parameter for a scattering angle of 95∘ . (b) What fraction of the particles will be scattered at angles greater than 95∘ ? (c) Compute the approximate radius of an iron atom.

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Answer: Approximate radius of an iron atom ≈ 5.77 × 10^-15m or 0.0577 nm

(a) Impact parameter refers to the perpendicular distance between the path of an incident particle and the center of the target.

Let’s compute the impact parameter using the formula below:

b = cot(θ/2) * (R1 + R2),

where θ = scattering

angle = 95°,

R1 = radius of incident particle,

R2 = radius of target particle

R1 = r, radius of alpha particle

R2 = R, radius of iron atom

R = 1.2 × 10^-10mb = cot(θ/2) * (R1 + R2)

b = cot(95°/2) * (1.2 × 10^-10m + 8.5 × 10^-15m)

b = 3.556 × 10^-15m ≈ 4 μm

Answer: Impact parameter = 4 μm

(b) Scattering occurs when the impact parameter is less than the distance of closest approach (DCA).

Therefore, we need to compute DCA using the formula below:

DCA = 2R(R1 + R2) / (R + R1)sin²(θ/2)

DCA = 2(1.2 × 10^-10m)(17 × 10^-6m + 1.2 × 10^-10m) / (8.5 × 10^-15m + 1.2 × 10^-10m)sin²(95°/2)

DCA = 3.28 × 10^-13m

To determine the fraction of particles scattered at angles greater than 95°,

we need to compute the area of the ring centered at the point of closest approach, but outside of the minimum scattering angle.

The area of this ring is πb² (1 - cos(θ/2)).

Also, the total area of scattering is πDCA² / 4.

Therefore, the fraction of particles scattered at angles greater than 95° is:Fraction = πb² (1 - cos(θ/2)) / πDCA² / 4

Fraction = (4b² / DCA²) (1 - cos(θ/2))

Fraction = (4(4 × 10^-12m²) / (3.28 × 10^-13m)²) (1 - cos(95°/2)) ≈ 0.038

Answer: Fraction of particles scattered at angles greater than 95° ≈ 0.038

(c) To compute the approximate radius of an iron atom, we need to use the formula for the radius of the nucleus of an atom. The Rutherford scattering formula, which is based on Coulomb's law, was used to determine the size of the nucleus of the atom.

Rutherford's formula states that the radius of the nucleus is R = R0A¹/³,

where

A is the mass number of the nucleus,

R0 is the radius of the nucleus of hydrogen (which is about 1.2 × 10^-15m), and

R is the radius of the nucleus of the atom.

Therefore, we need to determine the mass number of iron. The mass number of an atom is equal to the number of protons (atomic number) plus the number of neutrons. The atomic number of iron is 26, and the atomic mass is 56. Thus, the mass number is 56.

R = R0A¹/³R = (1.2 × 10^-15m)(56)¹/³ ≈ 5.77 × 10^-15m

Answer: Approximate radius of an iron atom ≈ 5.77 × 10^-15m or 0.0577 nm

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