Speed of fastest electron w\ workfunction & photon energy

[fredwacko40]

Hi please help me I have a test tomorrow worth 30% and I can't even work out the most basic of questions because my book assumes prior physics and knowing constants etc which I don't know!

The work function of tungsten is 4.50 eV. Calculate the speed of the fastest electrons ejected from a tungsten surface when light whose photon energy is 5.80 eV shines on the surface.

Ok so I've read all the theory and I haven't found anything under the photoelectric section converting KE to speed, I asked my friend and he said KE = (1/2)m*v^2 which didnt work. So far I have

hf = Kmax + (I) where (I) is that funny I in a O symbol
5.8 = Kmax + 4.5
Kmax = 1.3

Im assuming that's correct but please correct me if I am wrong, so I now have the max Kinetic energy but no way to convert it to speed. The back of the book says the mass of an electron is 9.109*10^-31 kg

I've also tried an equation I found on the internet:

V^2/C^2 = (2*Kmax)/mc^2 and after a little searching I found c to be the speed of light, but that returned a value of V = 68.81 which although a nice number wasnt right, 676 km/s is meant to be the answer, perhaps I just have to convert it? I think its in m/s atm though so that would make it 0.06881 km/s? That would be way off... please help me! I need to pass this test!

 

[Doc Al]

hf = Kmax + (I) where (I) is that funny I in a O symbol
5.8 = Kmax + 4.5
Kmax = 1.3

Convert the energy units from eV to Joules (the usual SI units). Then you can calculate the speed using the definition of KE.

 

[fredwacko40]

Ok so the conversion from
eV to joules is 16.02*10^-19
which gives me 2.0826*10^-19

So I plug in the numbers

((KE)/(0.5*m))^(0.5) = V
and that gives me a REALLY tiny number
which is correct but its too small!

6.76*10^-26

Is it because I am putting the mass in as Kg? Thats the first thing I am jumping to because although that's the way the book has the weight of an electron it strikes me as a silly unit for an electron!

Edit: What would be a better unit?

 

[Doc Al]

Ok so the conversion from
eV to joules is 16.02*10^-19
which gives me 2.0826*10^-19

OK.

So I plug in the numbers

((KE)/(0.5*m))^(0.5) = V
and that gives me a REALLY tiny number
which is correct but its too small!

Do this calculation over. Carefully. (Kg is the proper unit for mass; note that the mass of the electron is tiny.)

 

[fredwacko40]

Thankyou so much! I don't know what I put in wrong the first time but I now have the correct answer :) I probably would have spent another hour on that not even realising Id already had it right but put in the wrong numbers!