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

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 ? = 290 nm.

V =

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

N =

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 =

What is the cut off frequency for this metal?

fc =

Explanation / Answer

The energy of emitted electrons is h*c/? - ?; this is (4.136*10^-15)*(3.00*10^8)/(290*10^-9) - 2.70 eV = 1.5786 eV. The voltage needed is 1.5786 V

The energy of each photon is h*c/? = (6.626*10^-34)*(3*10^8)/(290*10^-9) = 6.8544*10^-19 J

1.20*10^-6 W = 1.20*10^-5 J/s

Photons per second = 1.20*10^-6 / 6.63*10^-19 = 1.8099*10^12

Photocurrent = photons/sec * charge/photon = 1.8099*10^12 * 1.60*10^-19 = 2.8959*10^-7 A

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

h*? = 2.7 eV

? = 2.70/(4.136*10^-15)

? = 6.528*10^14 Hz

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.