A certain metal has a work function of ? = 2.5 eV. It is illuminated in vacuum b

Question

A certain metal has a work function of ? = 2.5 eV. It is illuminated in vacuum by 1.1 x 10-6 W of light with a wavelength of ? = 230 nm.

1)

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

V =

Volts

2)

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

N =

photons per second

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 =

amperes

4)

What is the cut off frequency for this metal?

fc =

Hz

Explanation / Answer

a) The energy of emitted electrons is h*c/? - ?; this is

(4.136*10^-15)*(3.00*10^8)/(230*10^-9) - 2.5 eV = 2.89478 eV.

The voltage needed is 2.89 V

b) The energy of each photon is h*c/? = (6.626*10^-34)*(3*10^8)/(230*10^-9) = 8.64*10^-19 J

1.1*10^-6 W = 1.1*10^-5 J/s

Photons per second = 1.1*10^-6 / 8.64*10^-19 =1.27*10^12

c) Photocurrent = photons/sec * charge/photon = 1.27*10^12 * 1.60*10^-19 = 2.036*10^-7 A

d)Cutoff frequency (I interpret that as the lowest light frequency needed to emit electrons):

h*? = 2.5 eV

? = 2.5/(4.136*10^-15)

? = 6.044*10^14 Hz

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