The work function of a certain metal is 6.6 eV. When a photon whose frequency is 2.80 x 1015 Hz is absorbed by an electron below the surface, the electron escapes the surface with 3.5 eV of kinetic energy. How much energy (in eV) did the electron use to reach the surface?
a) 1.59
b) 4.22
c) 1.50
d) 3.62
None of these
Solution
We have the formula in the context of photoelectric effect,
Incident photon energy= Work function of metal + Kinetic energy + Energy required to move electron to the surface.
Hence we can calculate required quantity as,
Energy required to move electron to the surface= Incident photon energy- Work function - Kinetic energy.
Given work function=6.6eV
Kinetic energy= 3.5eV
Incident photon frequency, = 2.8*10^15 Hz
Using Plancks constant value h=6.626*10^-34Js and 1eV= 1.602*10^-19J,
Incident photon energy,Hence,
Energy required to move electron to the surface,
Eventhough option c)1.50eV is closest to the actual value, since there are no tolerance limits given to the final answer, correct answer is, none of these...
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