The stopping potential for photoelectrons from a metal is 2.0 V when light of wavelength 310 nm is used. The stopping potential, in volts, when light of wavelength 155 nm is used is (hc = 1240 eV nm)

Question

TestHub · Accepted Answer

Photon energy at 310 nm = 1240/310 = 4.0 eV, so φ = 4.0 - 2.0 = 2.0 eV.

Photon energy at 155 nm = 1240/155 = 8.0 eV.

K.E.max = 8.0 - 2.0 = 6.0 eV, so the stopping potential is 6.0 V.

Trap: the stopping potential is not proportional to the frequency (that would give 4 V).

Answer: 6.

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