Photons of energy 5 eV illuminate cathode C and eject photoelectrons of Kₘₐₓ = 2 eV. If incident photon energy becomes 6 eV, photoelectrons cannot reach anode A when its potential relative to C is:
Options

Question

Photons of energy 5 eV illuminate cathode C and eject photoelectrons of Kₘₐₓ = 2 eV. If incident photon energy becomes 6 eV, photoelectrons cannot reach anode A when its potential relative to C is:

Options

G1 Physics · Accepted Answer

Work function W = 5 − 2 = 3 eV. With 6 eV photons, Kₘₐₓ = 6 − 3 = 3 eV. Stopping requires an anode potential at or below −Kₘₐₓ/e = −3 V relative to cathode. Answer: D.