A certain metal has a work function of = 2.7 eV. It is illuminated in vacuum by

Question

A certain metal has a work function of = 2.7 eV. It is illuminated in vacuum by 1.2 x 10-6 W of light with a wavelength of = 280 nm.

1.What is the minimum voltage, V, needed to reduce the resulting photocurrent to zero?

V =

2.How many photons per second, N, are incident on the metal?

N =

3.For the photon flux in part 2, what is the maximum possible photocurrent in amperes? [Note that in reality one never sees this, as the likelihood for any photon to eject an electron -- even if the photon has enough energy -- is quite small.]

Imax =

4.What is the cut off frequency for this metal?

fc =

Explanation / Answer

= 2.7 eV
= 280 nm

(1)
The energy of emitted electrons = h*c/ -
E = (4.136*10^-15)*(3.0*10^8)/(280*10^-9) - 2.70 eV
E = 4.43 eV - 2.70 eV
E = 1.73 eV.
Therefore voltage needed is, V = 1.73 V

(2)
The energy of each photon = h*c/
E = (6.626*10^-34)*(3*10^8)/(280*10^-9)
E = 7.1*10^-19 J

Power = 1.20*10^-6 W

Photons per second, N = (1.20*10^-6) / (7.1*10^-19)
Photons per second, N = 1.69*10^12

(3)
Photocurrent = photons/sec * charge/photon
Imax = 1.69*10^12 * 1.60*10^-19
Imax = 2.704*10^-7 A

(4)
E=h*f,
where
f = frequency
h = Planck's constant = 4.135*10^-15 eV*s

For Cutoff Freq -
Now, = 2.70 eV
h*fc = 2.70 eV
fc = 2.70/(4.135*10^-15)
fc = 6.5*10^14 Hz

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