Work function of Sodium is 2.75eV. What will be KE of emitted electron when photon of energy 3.54eV is incident on the surface of sodium?
Answers
Answer: 0.79 eV
Explanation:
Work function of sodium, ∅ = 2.75 eV
The photon energy, hf = 3.54 eV
We know that in the photoelectric the photon energy incident on the surface of the metal is equal to the sum of the work function of the metal and the maximum kinetic energy of the electron.
Therefore,
hf = K.E.max(electron) + ∅
⇒ K.E.max(electron) = hf - ∅ = 3.54 - 2.75 = 0.79 eV
Thus, the kinetic energy of the emitted electron is 0.79 eV.
Answer:
By conservation of energy,
energy of incident photon = max. K.E of photoelectron + work function
hv = K + Wo
therefore, K = hv - Wo
= 2.54 - 3.75
= 0.79 eV
Explanation: