Another physics question

If only I had mark schemes.

The power of a laser beam is 80mW. Calculate the number of electrons emitted per second from the caesium plate assuming that only 7% of the incident photons interact with the surface electrons.

Energy = Power x time
Energy = 0.08 x 1 = 0.08J

I calculated the energy of 1 photon as 3.9 x 10^-19

Number of photons emitted per second
= 0.08/(3.9 x 10^-19) = 2.05 x 10¹⁷ photons per second

So number of electrons emitted per second
= 0.07 x 2.05 x 10¹⁷ = 1.44 x 10¹⁶

Is this correct?

Reply 1

Golden Maverick

4

Widowmaker

If only I had mark schemes.

The power of a laser beam is 80mW. Calculate the number of electrons emitted per second from the caesium plate assuming that only 7% of the incident photons interact with the surface electrons.

Energy = Power x time
Energy = 0.08 x 1 = 0.08J

I calculated the energy of 1 photon as 3.9 x 10^-19

Number of photons emitted per second
= 0.08/(3.9 x 10^-19) = 2.05 x 10¹⁷ photons per second

So number of electrons emitted per second
= 0.07 x 2.05 x 10¹⁷ = 1.44 x 10¹⁶

Is this correct?

Looks good.

Reply 2

CrazyChemist

hmm - does the question mention a work function? because with the photoelectric effect, to release one electron from the metal you need to supply energy equivalent to or greater than the work function yesno?

as far as i can tell, you have calculated the number of photons that are absorbed by the electrons in the metal per second. This would be the right answer if a single photon causes a single electron to be emmitted - which may be the case, I'm not so hot on this. Anyone care to clarify?

Reply 3

Christophicus

OP

18

CrazyChemist

hmm - does the question mention a work function? because with the photoelectric effect, to release one electron from the metal you need to supply energy equivalent to or greater than the work function yesno?

as far as i can tell, you have calculated the number of photons that are absorbed by the electrons in the metal per second. This would be the right answer if a single photon causes a single electron to be emmitted - which may be the case, I'm not so hot on this. Anyone care to clarify?

the energy of the photons is greater than the work function so each photon would cause an electron to leave the material. Is this right?

The work function of caesium (in the question) is 3.0 x 10^-19J
The energy of the photon (I calculated this) is 3.9 x 10^-19J

Reply 4

corkskrew

2

CrazyChemist

hmm - does the question mention a work function?

Widowmaker

I calculated the energy of 1 photon as 3.9 x 10^-19

looks like the work function has been taken care of, but indeed it does need to be taken into account. Waves have energy E=hf and the energy of the wave must exceed the work function.

Reply 5

CrazyChemist

Aha - i get it. Just to make sure: if the energy of the photons is less than the work function, no electrons will ever be emitted, and only one electron can be released by one photon - the rest of the energy goes into kinetic energy of the photon? that sounds right yes?

Reply 6

corkskrew

2

exactly. and this is proof of the particle like nature of light - specific quanta of energy in each wave. it's electron excitation - the same principle that causes emission and absorption spectra, but with greater energy and only one wavelength (spectra are caused by the light in the visible range). you just assume that the surplus energy is used for kinetic energy.