Answer on Question #58591, Physics / Quantum Mechanics
Find work function for a silver surface for which the threshold frequency of incident light is 1.1×1015Hz.
Find: A – ?
Given:
umin=1,1×1015Hz
h=6,626×10−34J×s
Solution:
Equation of external photoelectric effect:
hu=A+2mvmax2(1),
where hu – energy of photon,
A – electron work function of the metal surface,
2mvmax2 – the maximum kinetic energy of the electron
Of (1) ⇒humin=A(2)
Of (2) ⇒A=7,29×10−19J
1eV−1,6×10−19J
A −7,29×10−19J
A=4,56eV
Answer:
4,56eV
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