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# Photoelectric effect question watch

1. Use Einsteins photoelectric effect equation to a surface of work function 2.0x10^-19 J to calculate the frequency of radiation needed to eject electrons from the surface with a max kinetic energy of 3.0x10^-19 J

So i started off using Kemax=hf-thi but they use this equation hf=1/2mv^2max + thi we havnt even seen this equation before and it isnt in my textbook. How am i meant to know to use this and how is it derived?
2. (Original post by hazbaz)
Use Einsteins photoelectric effect equation to a surface of work function 2.0x10^-19 J to calculate the frequency of radiation needed to eject electrons from the surface with a max kinetic energy of 3.0x10^-19 J

So i started off using Kemax=hf-thi but they use this equation hf=1/2mv^2max + thi we havnt even seen this equation before and it isnt in my textbook. How am i meant to know to use this and how is it derived?
K.E max is = 1/2mv^2
They have subbed that in and then re-arranged to make hf the subject

If you dont know where 1/2mv^2 comes from then it is just Kinetic energy..
Energy= Work done = Force x Displacement
Force= Mass x Acceleration
Energy=Ma x S
V^2=u^2 + 2as
as= V^2-U^2/2

We assume initial velocity = 0
so a= V^2/2
Energy=V^2/2 x M
= 1/2mv^2
3. (Original post by Jian_Earle)
K.E max is = 1/2mv^2
They have subbed that in and then re-arranged to make hf the subject

If you dont know where 1/2mv^2 comes from then it is just Kinetic energy..
Energy= Work done = Force x Displacement
Force= Mass x Acceleration
Energy=Ma x S
V^2=u^2 + 2as
as= V^2-U^2/2

We assume initial velocity = 0
so a= V^2/2
Energy=V^2/2 x M
= 1/2mv^2
Ah right yeah silly me! thanks for a min there I was thinking i had missed something really important lol cheers dude.

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