How Does Momentum Transfer Affect Scattering Angles in Rutherford Scattering?

[JamesJames]

Calculate the Rutherford cross section for a 10 MeV alpha particle scattered at firstly an angle greater than 90 degrees and secondly for an angle greater than 10 degrees.

What does the relative magnitudes tell you in relation to the probabilities of the momentum transfer at large and small angles?

I cna calculate the cross section for those two values but what does the second part of the question mean? I am too confused about this part of the question. How does momentum transfer enter the Rutherford problem?

Any help would be greatly appreciated !

 

[Andrew Mason]

Calculate the Rutherford cross section for a 10 MeV alpha particle scattered at firstly an angle greater than 90 degrees and secondly for an angle greater than 10 degrees.

What does the relative magnitudes tell you in relation to the probabilities of the momentum transfer at large and small angles?

I cna calculate the cross section for those two values but what does the second part of the question mean? I am too confused about this part of the question. How does momentum transfer enter the Rutherford problem?

I don't really understand the question, because momentum transfer is a function of the deflection angle. Momentum transfer depends on the deflection angle. The larger the deflection angle, the more change in momentum of the alpha particle (which is the amount of momentum transferred to the nucleus).

It would make more sense if the question asked how the probability of momentum transfer can be deduced from the relative scattering cross sections for the large and small angles. The ratio of cross-sections for a deflection angle of more than 90 degrees to that for greater than 10 degrees corresponds to the probability of large momentum transfer.

AM

 

[wais]

Rutherford scattering is a fundamental phenomenon in nuclear physics that was first discovered by Ernest Rutherford in 1911. It involves the scattering of alpha particles (helium nuclei) by a target nucleus, which provides valuable insights into the structure and composition of the nucleus.

To calculate the Rutherford cross section for a 10 MeV alpha particle scattered at an angle greater than 90 degrees, we can use the following formula:

σ = (Z1 * Z2 * e^2) / (4 * π * ε0 * E^2 * sin^4(θ/2))

Where σ is the Rutherford cross section, Z1 and Z2 are the atomic numbers of the alpha particle and the target nucleus, e is the elementary charge, ε0 is the permittivity of free space, E is the kinetic energy of the alpha particle, and θ is the scattering angle.

Substituting the given values, we get:

σ = (2 * 79 * (1.602 * 10^-19)^2) / (4 * π * (8.854 * 10^-12) * (10 * 10^6)^2 * sin^4(90/2))

= 9.34 * 10^-22 m^2

Similarly, for an angle greater than 10 degrees, we get:

σ = (2 * 79 * (1.602 * 10^-19)^2) / (4 * π * (8.854 * 10^-12) * (10 * 10^6)^2 * sin^4(10/2))

= 1.06 * 10^-20 m^2

The relative magnitudes of these two values tell us that the probability of momentum transfer is higher at larger scattering angles. This is because at larger angles, the alpha particle experiences a larger deflection and therefore undergoes a greater change in momentum. This is also supported by the inverse relationship between the cross section and the scattering angle in the formula.

Momentum transfer is an important factor in the Rutherford problem as it helps us understand the interaction between the alpha particle and the target nucleus. The greater the momentum transfer, the stronger the interaction between the two particles. This allows us to study the composition and structure of the nucleus in more detail.