What is the work function of the metal hit by photons with wavelength of 450 nm?

[fluidistic]

Homework Statement

Photons with wavelength of 450 nm hit a metal. The trajectory of the most energetic electrons detached from the metal follow a circular path of radius 0.2 m due to a magnetic field of magnitude $2 \times 10 ^{-5}T$. What is the work function of the metal?

Homework Equations

Lorentz's force. Kinetic energy of electrons: $\gamma m_e v^2$.
Centripetal force: $F_c = \frac{m_e v^2}{r}$.
Conservation of energy.

The Attempt at a Solution

The energy of a photon hitting the metal is equal to the work function of the metal plus the kinetic energy of a most energetic electron.
$E=hc/lambda \approx 4.42 \times 10 ^{-19}J=2.76 eV$.
I reach, using the mentioned equations (and using the fact that the centripetal force is worth the Lorentz force, that is $F_c=e^- vB$) that v of the most energetic electrons is worth $v=\frac{e^- Br}{m_e}$. It seems that the gamma factor for my calculator is worth 1 so basically the speed of the most energetic electrons aren't big at all compared to c.
This gives me a kinetic energy of approximately $4.51 \times 10^{-19}J$ or 2.81 eV; which is greater than the energy of any photon, which is totally impossible.
For the arithmetics/algebra, I used as many digits as I could.
$m_ e=9.10938215 \times 10 ^{-31}kg$.
$e^- =1.602176487 \times 10 ^{-19}C$. However I took c as 3x10^8 m rather than the exact value, but this won't change anything anyhow.
$h=6.62606896 \times 10^{-34}J$.

What am I doing wrong?Edit: I just see that I forgot to divide by 2 the kinetic energy... I don't think it should change a lot, but still, I'll redo the math.

Edit 2: I get that the work function is worth approximately $1.350055399 eV$ which seems in theory possible. However when I look in tables on the Internet, I see no metal even close to this. The smallest work functions I see are over 2 eV.
Can someone confirm/infirm my answer? Thank you.

 

[kuruman]

You are correct, this is not a relativistic calculation. I assume that you found the correct photon energy to be 2.76 (I didn't do the calculation). Now, how do you find the speed v that goes into evB?

 

[fluidistic]

You are correct, this is not a relativistic calculation. I assume that you found the correct photon energy to be 2.76 (I didn't do the calculation). Now, how do you find the speed v that goes into evB?

Yes I am sure I've calculated well the energy of the photon. I just checked it using Mathematica and this confirms my previous answer.
About your question, what I had done is $F_c =e^- vB=\frac{v^2m_e}{r}\Rightarrow v^2-\frac{ve^- Br}{m_e}=0$ which gives v=0 (discarded) or $v=\frac{e^- Br}{m_e}$.

 

[kuruman]

I get the same number (1.35 eV) as you. It may be low, but it is what it is considering the numbers that are given to you.

 

[fluidistic]

I get the same number (1.35 eV) as you. It may be low, but it is what it is considering the numbers that are given to you.

Thank you very much. I think I'll slightly complain to my homeword assister in university. She said we have to construct an intuition from the assignated problems, but in this case it seems like an unreal problem. For instance the problem should have specified a greater radius and I think it would have been more "real".

 

[kuruman]

Thank you very much. I think I'll slightly complain to my homeword assister in university. She said we have to construct an intuition from the assignated problems, but in this case it seems like an unreal problem. For instance the problem should have specified a greater radius and I think it would have been more "real".

Well, let's see what intuition this problem teaches you. The number given for the strength of the magnetic field is of the order of magnitude of the Earth's field. So if you were to conduct a photoelectric experiment, you have a feeling now of the effects of the Earth's field on the trajectory of the ejected photoelectrons. Not much.

 

[fluidistic]

I just talked with a friend about this problem and he told me that the kinetic energy of the electron isn't $\gamma m_e v^2$ but $(\gamma -1)m_e c^2$.
Hmm I must reredo all the math.
Edit:My calculator shows now that the kinetic energy of the electron is 0J (...). The speed of the electron is 703528.0596 m/s.

 

[kuruman]

So the speed of the electron is of order 1/1000 of the speed of light. Most people would call this speed non-relativistic. Calculate γ and see how close it is to 1. I don't think that the relativistic correction will be able to change significantly the size of the work function that you get.