Approximation to an atom

Smythe

Below is from some notes I have.

In an atom, an electron is held in an electrostatic potential caused by the charge on the electron and the nucleus.
The infinite box can be thought of as an approximation to this. If this is the case, what would the energy be? An atom has a size of ~10^-10 m, so the lowest energy of an electron trapped in an atom is of the order;
E ~ [(1.05 x 10^-34)^2 x ∏^2] / [2 x 9.8 x 10^-31 x 10^-20] ~ 5 x 10^-18 J

For the above, has the following equation, or some other, been used?

E = [m^2 x (h-bar)^2 x ∏^2] / 2mL

Where it states 10^-10 m in the original question, I wonder did the author actually mean 10^-20 m, since this is what is used in the working out?
If the equation I have emboldened is the correct equation to use, they don't appear to have inserted the value of h-bar into the working out.

Thank you for any advice on this!

Morbert

Smythe wrote: »

Below is from some notes I have.

For the above, has the following equation, or some other, been used?

E = [m^2 x (h-bar)^2 x ∏^2] / 2mL

Where it states 10^-10 m in the original question, I wonder did the author actually mean 10^-20 m, since this is what is used in the working out?
If the equation I have emboldened is the correct equation to use, they don't appear to have inserted the value of h-bar into the working out.

Thank you for any advice on this!

The L should be squared in your equation:

E = [m^2 x (h-bar)^2 x ∏^2] / 2mL^2

Smythe

Thanks Morbert.

So since h-bar ~ 1.05x10^-34, and this value is in the top line of the calculation, are they therefore missing a value for m^2 in the top line?

citrus burst

The equation you give has one small mistake, the L should be squared. This would explain where the 10^-20.

http://en.wikipedia.org/wiki/Particle_in_a_box
http://ecee.colorado.edu/~bart/book/eband2.htm#well

Other then that everything should work out fine with the standard energy levels for an infinite potential well. I don't know why he omitted the calculation of h bar.

Morbert

Smythe wrote: »

Thanks Morbert.

So since h-bar ~ 1.05x10^-34, and this value is in the top line of the calculation, are they therefore missing a value for m^2 in the top line?

Since they are looking at the ground state, m on the top line is just 1.

They confusingly seem to be using m to denote both the quantum number, and also the electron mass.

Smythe

Thanks guys.

I understand it now.

I know what I'm doing wrong, I'm thinking about n^2 being on the top line, which in this case is 1^2 = 1 which is why it's not on the top line of the calculation.

Smythe

Also, in the quote in my first post we have;

E ~ [(1.05 x 10^-34)^2 x ∏^2] / [2 x 9.8 x 10^-31 x 10^-20] ~ 5 x 10^-18 J

Where there is the value of 9.8 x 10^-31, presumably this is wrong? The mass of an electron being 9.11 x 10^-31 kg.

Therefore the answer to the question becomes E ~ 6 x 10^-18

What do you think...?

Morbert

Smythe wrote: »

Also, in the quote in my first post we have;

E ~ [(1.05 x 10^-34)^2 x ∏^2] / [2 x 9.8 x 10^-31 x 10^-20] ~ 5 x 10^-18 J

Where there is the value of 9.8 x 10^-31, presumably this is wrong? The mass of an electron being 9.11 x 10^-31 kg.

Therefore the answer to the question becomes E ~ 6 x 10^-18

What do you think...?

Yes, I think 9.8 is a mistake.