AS Daily A Level Physics question
Electrons are accelerated from rest through a potential difference V and then diffracted by a thin graphite crystal, producing a pattern of concentric rings on a fluorescent screen. The accelerating potential difference V is then quadrupled. Which statement correctly describes the change to the radius of a specific ring in the diffraction pattern?
Answer
The correct answer is B.
The work done on an electron accelerated through a potential difference V is W = eV, where e is the elementary charge. This work done is converted into the kinetic energy (KE) of the electron, so KE = eV. The kinetic energy is also given by KE = p²/2m, where p is the momentum and m is the mass of the electron. Therefore, p² = 2meV, which means p is proportional to √V. The de Broglie wavelength (λ) is given by λ = h/p, where h is the Planck constant. This means that λ is inversely proportional to p, so λ ∝ 1/√V. In a diffraction experiment, the angle of diffraction for a particular feature (and thus the radius of the corresponding ring) is proportional to the wavelength (for small angles, sinθ ≈ θ ∝ λ). Therefore, the radius of the ring is proportional to 1/√V. If the potential difference V is quadrupled, √V is doubled. Since the radius is inversely proportional to √V, the new radius will be 1/2 of the original radius.