# Homework Help: Photoelectric Effect: Working out kinetic energy of photoelectron

1. Mar 17, 2014

1. The problem statement, all variables and given/known data
With the work function of copper being 4.65eV calculate the kinetic energy of a photoelectron knocked out with a photon with wavelength of 200nm . Then calculate the velocity of the electron.

2. Relevant equations

KE=hf-ø

3. The attempt at a solution
FIrst time I have done a question on this and what is confusing me is the KE comes out to be almost exactly equal to the work function but with a negative value, because hf is so small. This is what I have done

$$KE=hf-ø \\ KE=(6.621 \times 10^{-34})(\frac{3 \times 10^8}{2 \times 10^{-7}})-4.65 \\ KE=-4.649999999999999999006085 \\$$

I had to use wolfram alpha to get that result as every calculator I have used just rounded to -4.65, I dont know whether this is the whole point or not? If it is correct then I can just calculator the velocity with 1/2mv^2 and using the mass of the electron right?

Any help is appreciated,

thanks :)

Last edited: Mar 17, 2014
2. Mar 17, 2014

### ZapperZ

Staff Emeritus
Have you checked your units?! The work function is in electronvolts!

Carrying units throughout your calculation may be a pain, but it is your first "line of defense", and it tells you if you've done something silly. It also prevents you from being careless.

So keep the units for every single variable and constants that you use in the equation!

Zz.

3. Mar 17, 2014

Right, so it is 4.65eV, and 1ev is 1.6x10^(-19)J so that means the KE is = 4.65 x 1.6 x 10^(-19) = 7.44 x 10^(-19)J?

Or is it done like..
$$KE=hf-ø \\ KE=(6.621 \times 10^{-34})(\frac{3 \times 10^8}{2 \times 10^{-7}})-(4.65 \times 1.602 \times 10^{-19}) \\ KE=7.389 \times 10^{-37} J \\$$

Last edited: Mar 17, 2014
4. Mar 18, 2014

If it is how i did it the second time in the above post, the velocity then becomes...

$$KE=7.389 \times 10^{-37} J \\ \frac{1}{2} m_e v^2=7.389 \times 10^{-37} \\ v=\sqrt{\frac{7.389 \times 10^{-37}}{2m_e}} \\ v=\sqrt{\frac{7.389 \times 10^{-37}}{2(9.12 \times 10^{-31}}}=6.36 \times 10^{-4} m/s \\$$

or if its the first its
$$KE=7.4 \times 10^{-19} J \\ \frac{1}{2}m_ev^2=7.4 \times 10^{-19} \\ v=\sqrt{\frac{7.4 \times 10^{-19}}{2m_e}} \\ v=\sqrt{\frac{7.4 \times 10^{-19}}{2(9.12 \times 10^{-31}}}=6.4 \times 10^5 m/s =640km/s\\$$

A big difference, the first is more likely, however I always envisiged them moving a lot faster than that. I think both must be wrong.

5. Mar 19, 2014

I would really appreciate it, like loads, if someone could let me know if either of the ways I did it in my second d post (post 3 in the thread), were correct. The homework is due in in 3 hours, thanks :)

6. Mar 19, 2014

### Tanya Sharma

You should be getting KE = 2.49 x 10-19J

Last edited: Mar 19, 2014
7. Mar 19, 2014

With the values I used in the post I keep coming up with the values I got earlier, should I be using, or rather would it make it easier if I used the value stated for the plank constant in EV?

Its too late now as I handed it in but at least I can then understand where I went wrong. Thanks :)

8. Mar 19, 2014

### Tanya Sharma

What value are you getting for hf in Joules?

9. Mar 19, 2014

### nasu

You have a problem with the exponents.
What is $\frac{10^{-34} \times 10^8}{10^{-7}}$?

10. Mar 19, 2014

I think i was doing it on my calculator wrong as now I am back home and using wolfram alpha I am now getting the 2.49*10^(-19} value for the KE.

I know it depends on the calculator but what I was doing, to make it easier, instead of getting lost in parantheses I first just did 300000000/0.0000007 and then multiplied that answer by planks constant in Joules and then presses enter to get the value for hf and then subtracted (4.65*1.602*10^(-19)). I can see it was a calcualtor user error somwhere along the line.

would it be $\frac{10^{-34+8}}{10^{-7}}=\frac{10^{-26}}{10^{-7}}=10^{(-26)-(-7)}=10^{-19}$ ?

11. Mar 19, 2014

### nasu

Yes, this is it. :)

You don't need a calculator for this, do you?
Use it just to do the 6.6*3/2 (if you really need a calculator for this) and you will be less likely to make a mistake of many orders of magnitude. :)
One of the reasons for using scientific notation is that you can split the calculation into two parts. The exponential part separately and the (small) numbers in front of the exponential factors separately.

12. Mar 19, 2014