Estimating Diameter of Hydrogen Atom from Lyman Alpha Radiation

[ultimateguy]

Homework Statement

The Lyman alpha radiation for hydrogen (nf=2, ni=1) has a wavelength of 1216 angstroms. Using the simplified model of an infinite 1D potential well, derive an estimate of the diameter of the hydrogen atom. How does this value compare with twice the Bohr radius 2r=1.06anstroms?

Homework Equations

I imagine maybe De Broglie's relation, or the energy of a potential well.

The Attempt at a Solution

I have no idea how I can relate a potential well to the radiation given off by hydrogen. I realize that I'm supposed to work this out myself, but my exam is tomorrow morning and would appreciate if I could get the solution, or at least a hefty hint.

 

[Physics Monkey]

Hi ultimateguy, you know the energy levels of an infinite well, right? What frequency of light would be emitted if an electron made a transition from n=2 to n=1 in an infinite well of size L?

 

[ultimateguy]

The energy levels for an infinite well are $$E_{n}=\frac{n^2h^2}{8mL^2}$$ and the frequency is $$\omega=\frac{E}{\hbar}$$.

EDIT: Is the length of the potential well the same as the diameter of the atom?

 

[Physics Monkey]

The energy levels for an infinite well are $$E_{n}=\frac{n^2h^2}{8mL^2}$$ and the frequency is $$\omega=\frac{E}{\hbar}$$.

EDIT: Is the length of the potential well the same as the diameter of the atom?

First, the frequency for n=2 to n=1 is $$\omega = (E_2 - E_1)/\hbar$$, not just E.

Second, they clearly want you to approximate the atom with the potential well, so what do you think will correspond to the atom diameter?

 

[ultimateguy]

I think that the length of the well approximates the atom. For $$E_1$$ the wavelength $$\lambda=2L$$ and for $$E_2$$ that $$\lambda=L$$. Although this is the wavelength of the wavefunction and not the photon.

 

[Physics Monkey]

If the length of the well approximates the diameter of the atom then you should figure out what the length of the well is.

 

[ultimateguy]

I tried doing that already. I had $$E_2-E_1=\frac{3h^2}{8mL^2}$$. I then substituted into $$\omega=\frac{E_2-E_1}{\hbar}$$ to get $$\omega=\frac{3h\pi}{4mL^2}$$ and solving for L I get $$L=\sqrt{\frac{3h\lambda}{4m}}$$. I got 7.77x10^-21m using this.

 

[Physics Monkey]

You did your algebra wrong. For one thing you forget a factor of c, the speed of light.

 

[ultimateguy]

Don't know where c comes in.

 

[Physics Monkey]

What are the units of frequency? How did you relate angular frequency to wavelength? What are the units of wavelength? Notice anything? Your current expression does not have the right units, so there is no way you could get some number of meters from it.

 

[ultimateguy]

$$\omega=\frac{2c\pi}{\lambda}$$. Damn it I'm an idiot.

 

[ultimateguy]

My equation for length is $$L=\sqrt{\frac{3h\lambda}{8mc}}$$. If I use the proton mass for m I get 7.76x10^-12m. This isn't right, is it?

 

[Physics Monkey]

Nope. Why are you using the proton mass?

 

[ultimateguy]

I figured it was a hydrogen atom, has one proton. Now I see that it is an electron mass I need to use, and I get 3.325x10^-10m. I guess the electron mass is used because that is what's in a potential well. That must be right :P.