Answer: Delta_lambda = (h/m e c)(1 - cos theta).
- A Delta_lambda = (h/m<sub>e</sub> c)(1 - cos theta)
- B Delta_lambda = h/p, the de Broglie wavelength formula for a particle
- C Delta_lambda = hc/(eV), an expression mixing in the electronvolt unit
- D Delta_lambda = h/(m<sub>e</sub> c), the Compton wavelength constant with no angle term
Correct answer: A. Delta_lambda = (h/m<sub>e</sub> c)(1 - cos theta)
Explanation: Compton shift: Delta_lambda = (h/m<sub>e</sub> c)(1-cos theta) where theta is scattering angle. At theta=90°: Delta = h/m<sub>e</sub> c = 2.43 pm.
In the photoelectric setup, light striking the metal plate ejects electrons that cross the evacuated tube to the collector, producing a measurable current on the ammeter.
Concept context
Photoelectric effect, de Broglie waves, Bohr's model, atomic spectra.