De Broglie Waves

1. Jul 25, 2008

conniechiwa

1. The problem statement, all variables and given/known data
Atoms are on the order of one Angstrom (A) or 10-10m in size while the diameter of the nucleus is approximately one hundred thousand times smaller. Let's pretend we have a "classical" atom where both the atom and its nucleus have a defined position and size. (This it contrary to the laws of quantum mechanics which appear to be true, but it's not a bad approximation.) Our atom has an diameter of 1 A and a nucleus that's 1.1 X10-15m in diameter. What is kinetic energy of a an electron whose de Brogile wave length is equal to the size of this atom?

2. Relevant equations
r1= nsquared*r
r2=nsquared*r
1/λ = R (1/(n1 squared) - 1/(n2 squared))
λ=hc/E

3. The attempt at a solution
r1= nsquared*r
(1.1E-15 m / 2)=nsquared*(5.29E-11 m)
n1=0.00322

r2=nsquared*r
(10E-10 m / 2)=nsquared*(5.29E-11 m)
n2=0.9722

1/λ = R (1/(n1 squared) - 1/(n2 squared))
1/λ = (10973731.6 inverse meters)(1/0.00322 squared - 1/0.9722 squared)
λ = 9.4485E-13

λ=hc/E
9.4485E-13 = (6.63E-34 J*s)(3E8 m/s) / E
E=2.1051E-13 J = 1314042.476 eV

2. Jul 25, 2008

Staff: Mentor

Take the size of the atom to be = diameter of 1 A (10-10 m) = $\lambda$.

What is the expression for the de Broglie wavelength in terms of the particle's momentum?

What is the relationship between momentum and kinetic energy?

3. Jul 25, 2008

muppet

I think you're trying to use the Rydberg formula? I'm afraid that that's conceptually wrong. The Rydberg formula gives the difference between energy levels at set radii, and you can't treat the position of the nucleus as an energy level.
As it happens, I don't understand why you've been given the size of the nucleus. The key phrases as far as I can see are "De broglie wavelength" and "size of the atom". You know the size of the atom. You have a formula relating energy and de broglie wavelength...

4. Jul 26, 2008

conniechiwa

KE = momentum squared / 2m
I'm unsure what to plug in for m though.

5. Jul 26, 2008

muppet

Well, you could use the mass of an electron :tongue:
But if you read your own post again you'll find you don't have to...

6. Jul 26, 2008

conniechiwa

Oh okay I figured it out. Thanks a bunch!