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pop000 · Mar 4, 2011, 11:48 AM

it known that proton of Electromagnetic radiation Cause to an electron emission from Metal surface.
the velocity of the electron is 3.6*10^3km/h

A) how I Calculate the Wavelength of the electron?
B) and how many energy are Necessary to Get out electron from the surface?

thanks.

jcaron2 · Mar 4, 2011, 08:55 PM

A) I believe you can use the de Broglie relation:

$\lambda=\frac hp=\frac h{m \cdot v}$

where h is Planck's constant, m is the rest mass of the electron, and v is the velocity (in m/s, not km/h).

B) I'm not really sure what you're asking here. It sounds like you need to know the work function (or the Fermi energy or the ionization energy), which requires knowing exactly what kind of metal it is. Is there more information you haven't shared?

What kind of Chemistry class is this? It must be quite advanced. This stuff is more in the realm of solid state physics.

pop000 · Mar 5, 2011, 02:15 AM

hi.
thanks for answer.
for A I will try.
B)i sorry I not shared all the Information.
Question B) need to be like this: when the frequency radiation are low than 2.5*10^16Hz‏ so there are no any electron emission from the Metal surface. Now I asked to find how much energy are necessary to remove an electron from the surface.
all I know about the surface that it an Metal surface.
and yes this Chemistry start to be a little bit more advanced. :)
thanks.

jcaron2 · Mar 5, 2011, 09:08 AM

Aaah. So if the radiation frequency needs to be at least 2.5e16 Hz, then you simply need to find the energy of a particle at that frequency.

You can do that with an alternate version of the de Broglie relation:

$E=hf$

$E=(6.626 \times 10^{-34} \text{J\cdot s}) (2.5 \times 10^{16} \text{s^{-1}}) = 1.66 \times 10^{-17} \text{J}$

Or if you prefer the answer in electronvolts, rather than joules,

$1.66 \times 10^{-17} \text {J} \cdot \frac {1 \text {eV}}{1.602 \times 10^{-19} \text {J}}=103.4 \text {eV}$

Unknown008 · Mar 5, 2011, 09:19 AM

I'd too say that's it would fit into Physics, with them radiation and photoelectric effect.

pop000 · Mar 5, 2011, 09:44 AM

Yep that truth I will ask in the correct area next time :)

pop000 · Mar 5, 2011, 09:44 AM

OK I will try this.

Thanks for helping

jcaron2 · Mar 5, 2011, 09:06 PM

I think it fits in with chemistry as well. I was just remarking that it must be a very advanced course. I'm impressed. I never took a chemistry course so advanced. :)

jcaron2 · Mar 5, 2011, 09:17 PM

By the way, here's a quick brain-teaser for you guys.

The formula relating frequency to wavelength is well known:

$v=f \lambda$

We also know the kinetic energy of an object:

$E=\frac 1 2 mv^2$

And the momentum of an object:

$p=mv$

So now let's examine the two de Broglie equations:

$\lambda = \frac h p$

$E=hf$

Now let's do some substitutions:

$\lambda = \frac h p$

$\frac v f = \frac h {mv}$

$mv^2=hf$

$mv^2=E$

How can we have $E=mv^2$ and yet $E=\frac 1 2 mv^2$ at the same time!

pop000 · Mar 6, 2011, 02:56 AM

Well I want to stat to study in the Institute of Technology (is a kind of University) so I have to take some course before.
And is only get harder :)

pop000 · Mar 6, 2011, 02:59 AM

If you will ask me this in more 1-2 months I will know to answer :)
But now I only start to study this Level,

Unknown008 · Mar 6, 2011, 10:07 AM

Well, I think that it must have something to do with the differences quantum physics and 'classical' physics, where an electron's behaviour will diverge from a larger mass' behaviour.

jcaron2 · Mar 6, 2011, 12:51 PM

It's actually because $v=f \lambda$ gives you the phase velocity of the object ( $v_p=\frac \omega k$ ).

However, the kinetic energy $E=\frac 1 2 mv^2$ is based upon the group velocity ( $v_g=\frac {\partial \omega}{\partial k}$ ).

When talking about matter waves, the two are NOT equivalent!

pop000 · Mar 6, 2011, 01:33 PM

by the way is possible that the equation E=Mc^2 will be equal to 0?
I ask because I read that object even if it not move, so it still have energy because the mass. SO E still get any value.
but if you take the equation E=Mc^2 and c get the value 0 so all the equation need to be equal to 0.

where I wrong?

thanks for all the information that u shared with us.

Stratmando · Mar 6, 2011, 05:13 PM

This may help, the antenna wavlength formula is the only one I use. Don't know if it will help you?
http://www.crompton.com/wa3dsp/hamradio/antcalc.html
I originally was going to post the formula, this is easier.

jcaron2 · Mar 6, 2011, 05:55 PM

Pop, since c is a constant (around 3e8 m/s), E=mc^2 should never equal 0 for any object with non-zero mass.

Einstein's equation demonstrates the equivalence of mass and energy. Any object with finite rest mass automatically has finite energy content.

By the way, in the case of photons, which DO have zero rest mass but travel at the speed of light, the formula is still valid. The formula simply states that photons with certain relativistic energy (E=hf from de Broglie's equation) have an equivalent relativistic mass. If you could grab a photon and hold it in place, however, so that it's energy dropped to zero, it's equivalent mass would drop to zero as well.

The Wikipedia article on mass-energy equivalence does a much better job explaining this than I do.

pop000 · Mar 7, 2011, 02:50 AM

Oh OK thank you again.

pop000 · Mar 7, 2011, 01:59 PM

for one second I thought I found a mistake in Einstein's equation :-)

jcaron2 · Mar 7, 2011, 03:59 PM

:)

pop000 · Mar 12, 2011, 03:01 PM

if I want to know the wavelength of the photon that removed out the electron. So all I need to do is to use in this formula: ג=C/V
when we know that v=2.5*10^16 yes?
I am not sure about v if this is the real frequency, you can read in the beginning of this discussion to see if I correct or not.

and 1 more thing: when you told me to use E=h*v this mean to find the energy and energy mean photon yes?
thanks.