[Unknown008]

Hm... your notation is strange...

If you use ג=C/V, the ג is the wavelength, C is the velocity of the photon and V is the frequency of the photon.

E = hf

(or in your equation, your frequency is denoted by v)

gives the mean energy of the photon, yes.

[pop000]

C is the Speed ​​of light maybe is the same with the velocity of the photon?!
what I ask is how can I know what is v (frequency)of the photon that removed out the electron?
According to the information that I given here.
if I know v I can use this:×'=C/V to know the wavelength of the photon.


thanks.

[Unknown008]

Since we are considering the photos as light, yes, we take it as the speed of light.

Yes, that's the same v.

[pop000]

OK thank you :)

[jcaron2]

To make matters more confusing, in optics the frequency is usually denoted with the Greek letter nu (), which looks very much like the English letter vee (v), which usually denotes velocity (but not in optics, since the velocity is usually written in terms of the speed of light, c). :confused:

By the way, Pop, to write that character (nu) in this forum, you write the word math in brackets [] (to turn on math mode), followed by \nu, followed by /math in brackets [] (to turn off math mode).

[pop000]

Hi.

I think I need to use this formula to answer about my second question "how many energy are Necessary to Get out electron from the surface"


About the math mode I will try to use it.
Thanks.

[pop000]

Hi.
Can you please tell me what kind of formula did you used for calculated for electronvolts?

Thanks.

[Unknown008]

1 electron = 1.602 x 10^-19 C

Similarly,

1 eV = 1.602 x 10^-19 CV

And since W = QV, we have Joules = Charge x Voltage.

So,

1 eV = 1.602 x 10^-19 J

And jcaron only used proportions.

1.602 x 10^-19 J = 1 eV
1 J = 1/(1.602x10^-19) eV
1.66 x 10^-17 J = [1/(1.602x10^-19)] * 1.66 x 10^-17eV

[pop000]

Thank you :)