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# What am I doing wrong? photolelectric effect

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1. Metal surface is illuminated with light of wavelength 410nm, photoelectrons are emitted with a maximum kinetic energy of 2.1x10-19 joules.

Calculate the work function of the metal surface.

Energy of the photon = C/lambda = F
= 3x10^8 / 410x10-9 = 7.317x10^14 HZ
energy of photon = HF = 6.63x10-34 * 7.312x10^14 = 4.85x10^-19 joules.

To work out work function

HF = workfunction + max KE
workfunction = HF - max KE
= 4.85x10-19 - 2.1x10-19 = 2.75x10-19 joules.

This is wrong? the answer is supposed to be 1.61x10-19 joules so I don't know what I have done wrong, can anyone help?

Thanks
3. Looks right to me too. Are you doing a book question?
4. (Original post by Hazza616)
Looks right to me too. Are you doing a book question?
Yep, its the turning points AQA book

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