Q284 · Practice questionComplex unfamiliar7 marks
QUESTION 284 (7 marks)
Light illuminates a metal of work function . The most energetic emitted electrons enter a uniform magnetic field perpendicular to their velocity and follow circular paths of radius . Assume single-photon emission and nonrelativistic electrons. Use , , and .
a)[3 marks]
Determine the maximum electron speed and kinetic energy from the curvature.
b)[2 marks]
Determine the wavelength of the incident light.
c)[2 marks]
Predict what doubling the incident intensity at fixed wavelength does to the maximum electron-path radius and to the electron emission rate. Assume unchanged efficiency and no saturation.
WORKED SOLUTION
7 marksPractice marking scheme
ANSWER
(a) ; . (b) (about ). (c) Maximum radius unchanged; emission rate doubles.
Worked solution
(a) gives . Thus . Since , the nonrelativistic assumption is consistent.
(b) . Hence .
(c) Photon energy is fixed by wavelength, so maximum electron kinetic energy, speed and magnetic radius remain unchanged. Twice the intensity supplies twice as many photons per second over the same illuminated area, doubling the emission rate under the assumptions.
Equivalent physically justified methods, alternative valid sketches and consistent units accepted.
Accept sensible significant figures and consistent rounding from stated constants.
Part a: Equate magnetic and centripetal force.
Part a: Calculate the speed.
Part a: Calculate the maximum kinetic energy.
Part b: Use the photoelectric energy balance including work function.
Part b: Calculate the incident wavelength.
Part c: Explain unchanged maximum radius using unchanged maximum kinetic energy.
Part c: Predict twice the emission rate using increased photon rate.
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