Question

Determine the classical differential cross section for Rutherford scattering of alpha particles of energy \(5.00 \mathrm{MeV}\) projected at uranium atoms and scattered at an angle of \(35.0^{\circ}\) from the initial trajectory. Assume that both the target and the projectile atoms are pointlike.

Short answer

Using the given parameters and formulas, we can calculate the impact parameter, distance of closest approach, and classical differential cross section for Rutherford scattering of alpha particles of energy 5.00 MeV projected at uranium atoms and scattered at an angle of 35.0°. The result is: dσ/dΩ = ( ( (8.99 x 10^9 Nm^2/C^2) * (+2e) * (+92e) )^2 ) / ( 16 * ( (8.01 x 10^-13 J)^2 ) * (rmin^4) )

Step-by-step solution

Convert energy and find impact parameter

First, convert the energy of the alpha particles (5.00 MeV) to kinetic energy. The energy conversion factor is 1 MeV = 1.602 x 10^-13 J. Kinetic Energy = (Energy in MeV) * (Energy conversion factor) Kinetic Energy = (5.00 Mev) * (1.602 x 10^-13 J/MeV) Kinetic Energy = 8.01 x 10^-13 J The Rutherford scattering formula can be derived as: b = (k * q1 * q2) / (E * sin^2(θ/2)) Here, k is the Coulomb's constant, equal to 8.99 x 10^9 Nm^2/C^2. q1 and q2 are the charges of the alpha particle and uranium nucleus, respectively. The alpha particle has a charge of +2e, and the uranium nucleus has a charge of +92e, where e = 1.602 x 10^-19 C. θ is the scattering angle, which is given as 35.0°. Convert this angle to radians: θ (in radians) = (35.0°) * (π/180) = 0.6107 radians Now, we can calculate the impact parameter (b): b = ( (8.99 x 10^9 Nm^2/C^2) * (+2e) * (+92e) ) / ( (8.01 x 10^-13 J) * sin^2(0.6107/2) )

Calculate distance of closest approach

Using the impact parameter (b) and the scattering angle (θ), we can find the distance of closest approach (rmin) using the formula: rmin = b / sin(θ/2) rmin = b / sin(0.6107/2)

Calculate differential cross section

Now that we have the impact parameter (b) and the distance of closest approach (rmin), we can calculate the differential cross section (dσ/dΩ) for Rutherford scattering using the following formula: dσ/dΩ = ( (k * q1 * q2)^2 ) / ( 16 * (E^2) * (rmin^4) ) Substitute the values obtained in the previous steps to find the differential cross section: dσ/dΩ = ( ( (8.99 x 10^9 Nm^2/C^2) * (+2e) * (+92e) )^2 ) / ( 16 * ( (8.01 x 10^-13 J)^2 ) * (rmin^4) ) Calculate the value of dσ/dΩ to find the classical differential cross section for Rutherford scattering of alpha particles of energy 5.00 MeV projected at uranium atoms and scattered at an angle of 35.0° from the initial trajectory, assuming that both the target and the projectile atoms are pointlike.

Key concepts

Differential Cross Section

The differential cross section is a measure of the likelihood that a particle will scatter at a specific angle when it encounters another particle or target. It's important because it helps scientists predict the distribution of scattering angles in a particle collision process. In the context of Rutherford scattering, the differential cross section quantifies how likely it is for an alpha particle to deflect at a given angle after interacting with a heavier nucleus, like uranium in this case.

Imagine throwing a handful of marbles at a large sphere, representing the target nucleus. The way those marbles get deflected depends on how close each one gets to the sphere, akin to the impact parameter. The differential cross section is like a map that shows the density of scattered marbles at different angles away from the sphere. Mathematically, it's expressed as a ratio of the scattered particles per unit of the solid angle (dσ/dΩ), indicative of the angular distribution of scattering. In classical physics, for a pointlike projectile and target, this is calculated by the Rutherford formula, which incorporates fundamental constants, the charge of the particles, and the kinetic energy of the incoming particle.

Alpha Particle Scattering

Alpha particle scattering is at the heart of Rutherford's experiment, which revolutionized our understanding of the atom. It occurs when alpha particles, which are essentially helium nuclei consisting of two protons and two neutrons, are fired at a target and are deflected by the electrostatic forces from the target's nucleus. This phenomenon is not just a historical curiosity; it's still relevant today in research involving nuclear reactions, particle physics, and even in medical applications like cancer treatment through radiation therapy.

The scattering process involves the alpha particles coming in close proximity to the target nucleus, being subjected to a force due to their electric charges. This interaction bends the path of the alpha particle, causing it to scatter. The degree of bending can reveal information about the charge distribution within the nucleus. In the problem, a 5.00 MeV alpha particle is projected at a uranium atom. The classical physics approach simplifies this complex quantum process by treating the alpha particle and uranium nucleus as point charges and using the Coulomb force to predict the scattering pattern.

Classical Physics

Classical physics encompasses theories and laws of physics that predate the modern, quantum, and relativistic models. This traditional framework is grounded in concepts such as Newton's laws of motion and Maxwell's equations, which together describe the majority of macroscopic physical phenomena quite well. In the scenario of Rutherford scattering, classical physics elucidates the interaction between charged particles using the Coulomb force without invoking quantum mechanics.

In our textbook problem, we apply classical physics to describe and calculate the trajectory and scattering behavior of an alpha particle incident on a uranium atom. The formulas and principles like conservation of energy and momentum, derived from classical electrodynamic theory, allow students to determine the experimentally observable quantities like impact parameter, distance of closest approach, and as discussed, the differential cross section. These calculations provide valuable insights into the structure of atoms and the principles governing their interaction, which serve as the stepping stones to more advanced concepts in quantum physics.