# Light-wavelength calculation

1. Feb 8, 2005

### Kahsi

Hi.

How do I calculate this:

How long is the shortest light-wavelenght that can breake loose a elektron from a natrium surface?

2. Feb 8, 2005

3. Feb 8, 2005

### Kahsi

Thank you for the link, clive.
This is what I've got now:

$$eV_{0} = \frac{hc}{\lambda}-\phi$$
$$h = 6,6261 * 10^{-34}$$
$$c = 2,99792458*10^8$$
$$\phi =2,28eV$$

I need $$eV_0$$ to calculate $$\lambda$$. How do I get it?

And ofcourse I want the longest wavelength and not the shortest as I said before.

4. Feb 8, 2005

### Staff: Mentor

$$eV_0$$ is the maximum KE of the ejected electron; to find the minimum energy photon (longest wavelength) needed to eject an electron, set $$eV_0$$ to 0.

5. Feb 9, 2005

### Kahsi

Okay so we got

$$\frac{hc}{\lambda} = \phi$$
$$h = 6,6261*10^{-34}$$
$$c\approx 3*10^8$$
$$\phi = 2,28eV$$

This gives us

$$\lambda\approx 8,7*10^{-26}$$

Is this correct? Maybe i should turn eV to joule first?

Btw: Is the answer in meter? (I'm new to frequency)

Last edited: Feb 9, 2005
6. Feb 9, 2005

### Staff: Mentor

use standard units

You'd better use consistent units throughout. Since h = 6.625 x 10-34 J-s, you'd better convert eV to Joules!

The standard unit for length is the meter, so that equation for $\lambda$ will give you meters.

7. Feb 10, 2005

### Kahsi

Another question. If I'm lighting up a natriumsurface with a quicksilver-lamp electrons will "release". But which formulas should I use to calculate which energy the fastest electrons will get?

8. Feb 10, 2005

### Staff: Mentor

See post #4 in this thread. You have the formula.
(Photon energy) - (energy needed to eject an electron) = maximum KE of ejected electron