MODERN PHYSICS F/SCI..-WEBASSIGN(MULTI)
MODERN PHYSICS F/SCI..-WEBASSIGN(MULTI)
4th Edition
ISBN: 9780357644782
Author: Thornton
Publisher: CENGAGE L
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Chapter 4, Problem 51P

(a)

To determine

The fraction of the alpha particles that is scattered between 1.0° and 2.0°.

(a)

Expert Solution
Check Mark

Answer to Problem 51P

The fraction of the alpha particles that is scattered between 1.0° and 2.0° is 0.1176.

Explanation of Solution

Write the expression for number of alpha particles scattered.

    f=πnt(Z1Z2e28πε0K)2cot2(θ2)        (I)

Here, f is the number of alpha particles scattered, t is the thickness of foil with n atoms per unit volume, Z1e is charge on the particle, Z2e is the charge on scatterer, θ is the scattering angle and K is the kinetic energy.

Write the expression to find the frequency between  1.0° and 2.0°.

    f=f(1°)f(2°)        (II)

Conclusion:

Substitute 2 for Z1, 0.32 μm for t, 5.90×1028m3 for n , 79 for Z2, 1° for θ , 8MeV for K , 9×109 for 14πε0 and 1.6×1019 for e  in equation (I).

  f(1°)=(π(5.90×1028m3)(0.32 μm106m1μm)((2)(79)(9×109)(1.6×1019eVm)2(8×106eV))2cot2(12))=π(1.888×1022m)((2)(79)(1.44×109eVm)2(8×106eV))2cot2(0.5)=0.157

Substitute 2 for Z1, 0.32 μm for t, 5.90×1028m3 for n , 79 for Z2, 2° for θ , 8MeV for K , 9×109 for 14πε0 and 1.6×1019 for e  in equation (I).

  f(2°)=(π(5.90×1028m3)(0.32 μm106m1μm)((2)(79)(9×109)(1.6×1019eVm)2(8×106eV))2cot2(22))=π(1.888×1022m)((2)(79)(1.44×109eVm)2(8×106eV))2cot2(1)=0.0394

Substitute 0.157 for f(1°) and 0.0394 for f(2°) in equation (II).

    f=0.1570.0394=0.1176

Thus, the fraction of the alpha particles that is scattered between 1.0° and 2.0° is 0.1176.

(b)

To determine

The ratio of alpha particles scattered through angles greater than 1°  to the number scattered through angles greater than 10° and angles greater than 90°

(b)

Expert Solution
Check Mark

Answer to Problem 51P

The ratio of alpha particles scattered through angles greater than 1°  to the number scattered through angles greater than 10° is 100.5 and angles greater than 90° is 1.31×104.

Explanation of Solution

At two different angles, everything will remain same except the angle, so the ratio of the number of α  particles scattered will only depends on the scattering angle.

    f1f2=cot2(θ12)cot2(θ22)        (III)

Conclusion:

Substitute 1° for θ1 and 10° for θ2 in equation (III).

    f(1°)f(10°)=cot2(12)cot2(102)=100.5

Substitute 1° for θ1 and 90° for θ2 in equation (III).

    f(1°)f(90°)=cot2(12)cot2(902)=1.31×104

Thus, the ratio of alpha particles scattered through angles greater than 1°  to the number scattered through angles greater than 10° is 100.5 and angles greater than 90° is 1.31×104.

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