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# Phy4 jan 2005 Question 6 watch

Any help would be great.
2. (Original post by droid)
Any help would be great.
what syllabus u doin mate?
3. Edexcel GCE
4. (Original post by droid)
Edexcel GCE
ah rite, im not doin that, but if u typed out the question id be happy to have a go at it. there seems to be a lot of sylabus overlap
5. a) Intensity = power/area(4[pi]r^2)

b) the table:

Well we know energy of one photon = hc/landa, so substituting the two beams into it, and finding the ratio of Ea/Eb gives us 1:1.5 (or 2/3 in fraction).

Now this part requires more thought. N is the Number of photons. They have said that they both have the SAME intensity. We have just found out that beam A has 150% more energy per photon than beam B. So for the two beams to be of the same intensity, one of them has to have a great number of photons. Meaning that Beam B must have more photons than Beam A, to make them equal the same intensity.
So the ratio of the number of Na/Nb must be 1.5:1 which is 3/2 .

C) Work function is the minimum energy required to liberate an electron for the surface of a metal.

The photon energies in each beam:
Beam A = 6.67x10^-19 OR 4.17eV
Beam B = 4.45x10^-19 OR 2.78eV

So looking at the energies, we can see that Beam B can only liberate electrons from Potassium.
And Beam A can liberate electrons from Potassium and Magnesium.

The question asks which metal does ONLY beam A emmite electrons from, which is magenesium.
6. (Original post by SinghFello)
a) Intensity = power/area(4[pi]r^2)

b) the table:

Well we know energy of one photon = hc/landa, so substituting the two beams into it, and finding the ratio of Ea/Eb gives us 1:1.5 (or 2/3 in fraction).

Now this part requires more thought. N is the Number of photons. They have said that they both have the SAME intensity. We have just found out that beam A has 150% more energy per photon than beam B. So for the two beams to be of the same intensity, one of them has to have a great number of photons. Meaning that Beam B must have more photons than Beam A, to make them equal the same intensity.
So the ratio of the number of Na/Nb must be 1.5:1 which is 3/2 .

C) Work function is the minimum energy required to liberate an electron for the surface of a metal.

The photon energies in each beam:
Beam A = 6.67x10^-19 OR 4.17eV
Beam B = 4.45x10^-19 OR 2.78eV

So looking at the energies, we can see that Beam B can only liberate electrons from Potassium.
And Beam A can liberate electrons from Potassium and Magnesium.

The question asks which metal does ONLY beam A emmite electrons from, which is magenesium.
I get the same except for in part b) I get the ratios the other way round. i.e. I get the first one to be 3/2 and the second to be 2/3 ...
7. (Original post by sonja)
I get the same except for in part b) I get the ratios the other way round. i.e. I get the first one to be 3/2 and the second to be 2/3 ...
you are correct

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