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A metal has work function $2$ eV. Incident light of frequency $1\times 10^{15}$ Hz gives max KE (using $h = 4.14\times 10^{-15}$ eV s):
A$2.14$ eV, since $h\nu = 4.14$ eV, $K_{max} = 4.14 - 2$
B$4.14$ eV, equal to the full incident photon energy
C$2$ eV, equal to the metal's work function value
DZero, since the frequency is below the threshold here
Answer & Solution
Correct answer: A. $2.14$ eV, since $h\nu = 4.14$ eV, $K_{max} = 4.14 - 2$
$h\nu = 4.14$ eV; $K_{max} = h\nu - \phi = 4.14 - 2 = 2.14$ eV.
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