Quantum

secretliker SG

The memory of a computer element can be erased by exposing it to ultra-violet radiation of wavelength 360nm for 20 minutes. The memory is contained in an insulated silicon film of exposed area 1.5 X 10^-9 m-2, and the intensity of the UV light of 20Wm-2.

Calculate

(i) The number of photons incident on the film in 1 second

(ii) The charge acquired by the film in 20 minutes of exposure, if 1% of the incident photons cause photoemission of electrons.

Explain why erasure would be slower if the memory were exposed to sunlight of the same total intensity.

eagle SG

i) Assume UV ray is shone directly onto the film, and not at an angle

Energy of photon = h*c/wavelength = 5.525e-19 J
Intensity is 20Wm-2, so for the exposed area, power is 3e-8 W

Thus, number of photons incident on film in 1 second = 3e-8 / 5.525e-19 = 5.43e10 photons

ii) 1 photon max 1 electron emission

so 1% means 5.43e8 electrons
electron charge = 1.6e-19 C

So total charge in 20 mins = 5.43e8 * 1.6e-19 * (20 *60 s) = 1.043e-7 C

The photon must have the exact energy/frequency (E=hf) to be able to cause the photoemission of exactly one electron. This means that only UV photons in this case has the right amount of energy to cause photoemission.

Since sunlight consist of various wavelengths of light, UV rays are only a fraction of sunlight (and thus fraction of the intensity). Hence, the amount of power from UV rays from sunlight will be lower than the original intensity of the UV ray in the first part of this question. This results in a slower erasure rate since less photoemission will result.

secretliker SG

Perfect answer. Thanks!