# Photoelectric effect calculations

1. Oct 3, 2011

### NewtonianAlch

1. The problem statement, all variables and given/known data

Light with wavelength 452 nm illuminates a surface. The work function of the surface is 5.77 electron volts.

1. What is the frequency of light hitting the surface?

2. What is the maximum kinetic energy of the electron ejected from this surface?

3. What backing voltage would need to be applied to stop all electrons ejected from this surface from reaching the anode?

3. The attempt at a solution

1) f = $\phi$/h

f = 5.77/(4.14*10^-15) = 1.39*10^15 Hz

2) K = (hc/$\lambda$) - $\phi$

K = (4.14*10^-15*3*10^8)/(452*10^-9) - 5.77 = -3.02eV

3) I'm not too sure how to start 3.

I'm fairly sure part 1 is correct, I didn't expect to get a negative value for part 2 so I'm sure that's not right, what did I do wrong?

2. Oct 3, 2011

### mstud

I disagree on part 1:

The speed of light is $3.00 \cdot 10^8 m/s$ and $c=f \lambda$ this gives $f=\frac {c}{\lambda} = \frac {3.00 \cdot 10^8 m/s}{452 \cdot 10^{-9} m}=6.64 \cdot 10^{14} Hz$.

What are the \phi and h you are using?

BTW, When using formulas, you should not delete heading 2. It is much easier to help when we can see which formulas and variables you've tried to use.

For example, I'm used to that h is used for Planck's constant h=6.63 * 10^34 Js, you have an h standing for something else.
Thus it is sort of difficult for me to see what you have thought here ...

3. Oct 3, 2011

### NewtonianAlch

Sorry about that, for h I used a value of 4.14*10^-15 eV.s - which was given in my book as equal to 6.63*10^-34 Js

So I figured I'd use that to deal with eV.

The value of phi is the value of 5.77eV, the given value in the question.

I think you're right, don't know why I used that weird formula to calculate the frequency, I will take a closer look at it now.

4. Oct 3, 2011

### mstud

Oh, so you calculated the frequency of the "electron radiation" leaving the surface...

Since 1) asked for the freqecy of the light hitting the surface you'd better leave the electron volts for a moment when answering it ... There the frequency depends on the wavelength of the incoming light, not the eV value from the emission, step 2 in the process.

5. Oct 3, 2011

### NewtonianAlch

I can't believe the silly error! Thanks for pointing that out =)

6. Oct 3, 2011

### mstud

If the value for Kmax is negative, the energy of the incoming light is not high enough for electrons to be emitted. Could this be the point here? However normally you won't then be asked question 3), but I still assume this is a possible conclusion.

Otherwise the given work function is too high, and there is an error in the problem.

2)
Kmax is given by $hf-\phi$, where f is the frequency of the incident photon. Using the frequency found in 1), which gives $4.14 \cdot 10^{-15} \cdot 6.64 \cdot 10^{14} - 5.77 eV =$, gives me the same negative value for Kmax as you found, which would imply there's no electron emission at all, because the incoming light has to little energy...

This is also (approximately) supported by:

$Kmax=4.14 \cdot 10^{-15} (6.64 \cdot 10^{14} -1.39 \cdot 10^{15})= -3.01 eV$ where the last value is the frequency you found for electron emission.

3) $Kmax=eV_0$, where e is the electron's charge (1.60217656535×10^{−19} coulombs) and V0 the stopping potential (voltage). This can be used to calculate the stopping voltage to be applied. Just divide Kmax by e...

7. Oct 3, 2011

### NewtonianAlch

It did plague me for a while, but discussing it with a few friends I found out that they had similar answers for part 2, but I didn't understand why. I can see what the negative result means from your explanation, thanks for that!

8. Oct 4, 2011