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AS Physics - Photoelectric Effect Help Please

1) A modern 200 Watt sodium street lamp emits yellow light of wavelength 0.6 micrometers. Assuming it is 25 per cent efficient in converting electrical energy to light, calculate the number of photons of yellow light per second.

2) A certain metal plate shows photoelectric emission of electrons for radiation up to a wavelength of 600 nm. If light of wavelength 500 nm is used, what will the maximum velocity with which electrons are emitted?

These are the formula's I was given for question 2:
c = 3 X 10 8 m/s h = 6.6 X 10 -34 Js mass of electron = 9.1 X 10 -31 kg



Please can someone show me in steps how to do this question as I don't have a clue. Thanks.
Reply 1
Avatar for laeti
laeti
1st part: if you know the efficiency, you can calculate the output power. Then use the relation between energy and frequency (E = hf) to find the energy of one photon, and the definition of power (energy per unit time) to find the number of photons per unit time.

2nd part: if you now the wavelength at which photoemission stops, you can calculate the work function.
(edited 13 years ago)
Reply 2
Avatar for soup
soup
11
1. Find total energy and divide by energy of one particle.
Use E = P x t and E = hf remembering lamp is 25% efficient
2. Use Ek=hf−hf0 E_k = hf - hf_0
Reply 3
Avatar for Luis456
Luis456
OP
Original post by soup
1. Find total energy and divide by energy of one particle.
Use E = P x t and E = hf remembering lamp is 25% efficient
2. Use Ek=hf−hf0 E_k = hf - hf_0


could you show me how to do question 1 please
Reply 4
Avatar for soup
soup
11
Original post by Luis456
could you show me how to do question 1 please


Well 200W means 200J/s we have one second so 200J
But lamp is 25% efficient so it is 25% of 200 = 50J

Energy of one photon = hf = h x 0.6 mircometers.

No. of photons = total energy / energy of one
Reply 5
Avatar for laeti
laeti
Original post by soup
Well 200W means 200J/s we have one second so 200J
But lamp is 25% efficient so it is 25% of 200 = 50J

Energy of one photon = hf = h x 0.6 mircometers.

No. of photons = total energy / energy of one


actually it's the wavelength that is 0.6 micrometres, not the frequency

you need c = f*lambda
Reply 6
Avatar for Luis456
Luis456
OP
Original post by soup
Well 200W means 200J/s we have one second so 200J
But lamp is 25% efficient so it is 25% of 200 = 50J

Energy of one photon = hf = h x 0.6 mircometers.

No. of photons = total energy / energy of one


Thanks :smile:
Reply 7
Avatar for soup
soup
11
Original post by laeti
actually it's the wavelength that is 0.6 micrometres, not the frequency

you need c = f*lambda


ah good point - totally misread
Reply 8
Avatar for Luis456
Luis456
OP
Original post by soup
ah good point - totally misread


er so is the working out u did wrong then?
Reply 9
Avatar for soup
soup
11
Original post by Luis456
er so is the working out u did wrong then?


Only for the energy of a photon which would now be = (hc)/0.6 micrometers

And you have to change final division aswell
Reply 10
Avatar for Luis456
Luis456
OP
Original post by soup
Only for the energy of a photon which would now be = (hc)/0.6 micrometers

And you have to change final division aswell


orite and could you show me in steps how to tackle the second question
Reply 11
Avatar for hazbaz
hazbaz
11
Original post by Luis456
orite and could you show me in steps how to tackle the second question


We arnt going to do your homework for you
Original post by hazbaz
We arnt going to do your homework for you


Thats rude-We used to do the same thing::colondollar:
Reply 13
Avatar for bijesh12
bijesh12
13
What is the numerical value of Planck's Constant ?
Reply 14
Avatar for soup
soup
11
Original post by bijesh12
What is the numerical value of Planck's Constant ?


If you look at the original post it is there :biggrin:

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