A Caesium surface (work function = 2.14 eV) is illuminated with 210 nm light fro

Question

A Caesium surface (work function = 2.14 eV) is illuminated with 210 nm light from a 1 mW laser.

a) How many photons per second are emitted by the laser? __________________ photons/s

Only a small fraction of these photons can interact with electrons at the surface. Suppose this fractional efficiency is 0.0023%. What is the current in the detector assuming all the emitted electrons are captured? ______________ nA

Note that current is charge(in Coulombs) per second. The charge of one electron is 1.6×1019C

Explanation / Answer

A.)

Energy = (h * c) /
h = 6.63 * 10^-34 J * s
c = 3 * 10^8 m/s
= 210 nm = 210 * 10^-9 m

Energy = (6.63 * 10^-34 * 3 * 10^8) ÷ 210 * 10^-9
Energy of one photon = 9.47 * 10^-19 J

Power = Energy/time
1 Watt = 1 Joule per second

1.00 mW = 1 * 10^-3 W = 1 * 10^-3 Joule per second

A 1.00 mW laser produces 1 * 10^-3 Joule per second
One photon = 9.47 * 10^-19 Joules

1 * 10^-3 Joule/second / 9.47 * 10^-19 Joules/photon = 1.055 * 10^15 photons/second

B.)

curent (I) = efficeieny * no of photon* charge on photon

Current = {0.023/100}*(1.05*10^15)*(1.6*10^-19) = 38.64 nA

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