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# A2 Physics A Turning points: photo electricity question

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1. How do I do (b) and (c)? Topic 2.3 Turning points (OPTION D)

2. When a metal surface is illuminated with light of wavelength 410 nm, photoelectrons are emitted with amaximum kinetic energy of 2.10 × 10−19 J.Calculate:
a the energy of a photon of this wavelength
b the work function of the metal surface
c the stopping potential for this surface illuminated by light of wavelength 410 nm

So for part
(a) E=hc/lambda =(6.63*10^-34)*(3*10^8)/410*10^-9
E=4.85*10^-19 j

then I thought
(b) work function = hf - Ek
=4.85*10^-19 - 2.10*10^-19
= 2.75*10^-19

but this is wrong textbook answer is 1.61*10^-19j

What am I doing wrong?
2. help with unit 2 physics AQA?
3. (Original post by smitj893)
help with unit 2 physics AQA?
You are going to want to go here http://www.thestudentroom.co.uk/forumdisplay.php?f=131 and press "post new thread" then ask your question
4. (Original post by Music With Rocks)
How do I do (b) and (c)? Topic 2.3 Turning points (OPTION D)

2. When a metal surface is illuminated with light of wavelength 410 nm, photoelectrons are emitted with amaximum kinetic energy of 2.10 × 10−19 J.Calculate:
a the energy of a photon of this wavelength
b the work function of the metal surface
c the stopping potential for this surface illuminated by light of wavelength 410 nm

So for part
(a) E=hc/lambda =(6.63*10^-34)*(3*10^8)/410*10^-9
E=4.85*10^-19 j

then I thought
(b) work function = hf - Ek
=4.85*10^-19 - 2.10*10^-19
= 2.75*10^-19

but this is wrong textbook answer is 1.61*10^-19j

What am I doing wrong?
I don't think you are doing anything wrong in part b. I think the textbook answer is wrong. I have the same book and it seems to be an error,

Part (c). The stopping potential is when eV = Ek max

So 2.1x10-19 / e = V = 1.3125 V
5. (Original post by SirRaza97)
I don't think you are doing anything wrong in part b. I think the textbook answer is wrong. I have the same book and it seems to be an error,

Part (c). The stopping potential is when eV = Ek max

So 2.1x10-19 / e = V = 1.3125 V
Ah thank you very much, that question was stressing me out :P annoying how often the textbook answers are wrong.

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