Question

Consider an experiment in which slow neutrons of momentum ¯hk are scattered by a diatomic molecule; suppose that the molecule is aligned along the y axis, with one atom at y = b and the other at y = −b. The beam of neutrons is directed in the zb direction. Assume the atoms to be infinitely heavy so that they remain fixed throughout the experiment. The potential due to the atoms as seen by the neutrons can be represented by a delta function, so: V (~r) = a[δ(x)δ(y − b)δ(z) + δ(x)δ(y + b)δ(z)] (a) Calculate the scattering amplitude, and differential cross section, in the first order Born approximation. (b) In what ways does the quantum result differ from what one would expect classically?

Answer 1

Check the explanation

Step-by-step explanation:

When we have an object in periodic motion, the amplitude will be the maximum displacement from equilibrium. Take for example, when there’s a back and forth movement of a pendulum through its equilibrium point (straight down), then swings to a highest distance away from the center. This distance will be represented as the amplitude, A. The full range of the pendulum has a magnitude of 2A.

position = amplitude x sine function(angular frequency x time + phase difference)

x = A sin(ωt + ϕ)

x = displacement (m)

A = amplitude (m)

ω = angular frequency (radians/s)

t = time (s)

ϕ = phase shift (radians)

Kindly check the attached image below to see the step by step explanation to the question above.