The Photoelectric Effect.

In the spirit of Albert Einstein [again] i'm going to do a post on his, well what should be - most famous discovery. 

In 1905 Albert Einstein published a paper on the Photoelectric Effect [which I spoke about on Wednesday, it was in his 'Annus Mirabilis' year]. The photoelectric effect is the effect in which delocalised electrons are 'emitted' from a metal, due to the absorption of light.

Albert Einstein concluded that light is a particle, called a photon. Yet, it is also a wave. This is particle-wave duality, in light. 

These delocalised electrons can only be 'freed' if the wavelength of the light is high enough to over come to work function, i'm going to call the work function W [it is sometimes noted as Wo] - W is the amount of energy need to free an electron.

If we have red light, at a wave length of 700nm hitting sodium, will electrons be released?

In any equation you will be given W, which for sodium is 3.6x10^-19 J. 

You are given the wave length too, 700 nm, or 7x10^-7 m - you will also be told h, which is Planck's constant, or 6.626x10^-34.

If you didn't know, the energy of a photon is equal to E=hf or E=hc/λ.

So, E = (6.626x10^-34 * 3x10^8)/7x10^-7 - this equals... 

2.84x10^-19 J!

So, like Einstein found out, red light does not excite electrons enough to let them 'escape' the sodium, as it is not higher than W - or 3.6x10^-19 J.

Now onto violet light, with a wavelength of 400nm! Once again lets do our equation, but this time we'll use 4x10^-7m instead of 7.

So, E = (6.626x10^-34 * 3x10^8)/4x10^-7 - this equals... 

4.97x10^-19 J!

This is high enough to allow electrons to escape, and with a little extra math, we can work out their velocity too!

We'll use the equation: hf=W+1/2mv^2. Where hf is what we've just done, W is the work function, m is the mass of an electron and v is the velocity.

Since we know hf (we've just worked it out) as 4.97x10^-19 J, then...

4.97x10^-19 J = W+1/2mv^2.

Substitute an electrons mass and the work function of sodium into it and we get: 4.97x10^-19 J = (3.6x10^-19)+1/2(9.11x10^-31)*v^2.

You will always be given the mass of an electron, just incase theirs any confusion.

We can then work out the velocity by subtracting things from both sides.

Minus W off both sides and we get 1.37x10^-19 (4.97-3.6).

So, 1.37x10^-19 = 1/2(9.11x10^-31)*v^2.

If we times both sides by two, then divide by 9.11x10^-31 we get...

3x10^-11. The equation now states 3x10^-11 = v^2.

If we square root both sides we get the velocity, which is...

5.48x10^5 m/s - thats is incredibly fast!

Sorry for the long winded and mathematically challenging post, when you read this i'll be in Scotland somewhere so thank you for reading, once again;

Thanks for reading, Ben.

(If you're wondering why this is Ben's post since I said he was away this was saved in the drafts - as he said above he's in Scotland, thanks)