Physics help!!!

A 50.0 W laser produces a 4.00 μs pulse of photons with wavelength 2.00 x 10–7 m. The pulse is used to partially discharge a negatively charged metal plate by the photoelectric effect.

Assuming that 90% of the photons cause an electron to escape, calculate the loss of charge from the plate.

Thanks in advance!!

(edited 3 years ago)

Reply 1

yankeecandle

Not sure.. but

(3x10^8)/(2x10^07) = 1.5x10^15 = frequency of pulse of photons
(1.5x10^15)x(4x10^6) = 6x10^9 photons
only 90% of these cause an electron to escape so 0.9 x 6x10^9 = 5.4x10^9 photons
(5.4x10^9)x(1.6x10^-19) = 8.64x10^-10 C

this could be very wrong lol??? i didnt use the planck constant so i probably went wrong somewhere

Reply 2

Mr Wednesday

Original post by izzyrose23

this could be very wrong lol??? i didnt use the planck constant so i probably went wrong somewhere

Yup, Plank's constant definitly required for this, 2nd line of calculation broken.

If you know laser power and pulse duration then that gives you total optical energy. You then need to divide by the energy per photon to get number of photons and that uses E=hf for E the photon energy, h = Plank's const, f= frequency.

Reply 3

po.01

Original post by Mr Wednesday

Thank you so much!! I have got the number of 90% of the photons but how do you calculate the loss of charge from that ? Sorry if that's a really stupid question - feel like I'm over-complicating all this.

Reply 4

nzy

Original post by po.01

If each of the photons cause a singe electron leave the surface of the plate, then the total loss of charge would be the sum of the charges of that same number of electrons, I'm assuming in coulombs.

Reply 5

po.01

Original post by nzy

If each of the photons cause a singe electron leave the surface of the plate, then the total loss of charge would be the sum of the charges of that same number of electrons, I'm assuming in coulombs.

Ohhh that would make sense - thank you !!

Reply 6

Mr Wednesday

Original post by po.01

Ohhh that would make sense - thank you !!

This is one of the key ideas behind quantisation of energy. Light carries energy as well defined packets and you need enough energy in ONE packet to do the job of liberating an electron from a surface. In practice, not all photons do the job, sometimes the energy gets “lost” in a different process, or the electron gets pulled back to the surface. That is quantified by something called “quantum efficiency”, and it’s worked into the question here with that “90% of photons liberate an electron” part.

A bit of a fun warning for the future ..... :smile:

. Lots of the physics you learn at GCSE / A level is a bit of a fudge or simplification, there are often special cases where things are a bit different. For very bright light sources (e.g. a focused high power laser) it’s possible to get so called “multi-photon physics” where two or more photons can interact with one “thing”, but that’s university level stuff. That process has lots of real world applications, for example green laser pointers are not actually “green lasers”, they use an infra-red laser to make lots of invisible photons, then use a crystal to capture and add together pairs of photons to make green light.