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a. The threshold frequency of the surface.
f0=c/?0=[3.0x108m/s]/[0.45x10-6m]
=6.67x1014Hz.
b. The workfunction of the surface in eV.
W0=hf0=[6.63x10-9x6.67x1014]/[1.602x10-19]=2.76eV
c. The minimum speed with which a photoelectron is emitted if the frequency of the radiation is 7.5x1014Hz.
6.63x10-9x7.5x1014=[ 6.63x10-9x6.67x1014]+[½x9.11x10-6x ?2]
?=[12.081x1010]½ =3.4754x105m/s.
Moraa orina answered the question on April 12, 2018 at 07:14
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