# Homework Help: Particle in an infinite potential well

1. Nov 13, 2005

### jdstokes

Hi all,

I have an exam tomorrow and this problem concerns me greatly.

An electron is located in an infinitely deep one-dimensional square potential well. The width of the well is 1.00 nm.

(e) Light is shone on the electron causing it to jump from the ground state to the n = 3 state. What is the wavelength of the light?

(f) What would happen if light with wavelength twice that calculated in part f were shone on the electron in its ground state.

This does not correspond to a transition energy from the ground state to a higher energy state and so the photons will not be absorbed.

I'll work through the problem algebraically to show why I think the photon will in fact be absorbed.

The energy of the nth state is $E_n = n^2E_1$ where $E_1 = \frac{h^2}{8mL^2}$ is the energy of the ground state.

The transition energy between n = 3 and and the ground state is thus

$\Delta E = E_3 - E_1 = 3^2E_1 - E_1 = 8E_1 = \frac{h^2}{mL^2}$.

So the wavelength of the incident photon is

$\lambda = \frac{hc}{\Delta E}$.

If the light in part (f) has twice this wavelength, then the energy of the photons is half the value in part (e), ie

$\Delta E' = 1/2\Delta E = 4E_1$,

Since the transition energy between the ground state and the n = 2 state is just $3E_1$, why won't the electron be promoted to the n = 2 state?

Thanks.

James

2. Nov 13, 2005

### SpaceTiger

Staff Emeritus
Let's say it were. What happens to the remaining E1 of energy that was in the photon?

3. Nov 13, 2005

### jdstokes

Maybe it is emitted in the form of another photon.

4. Nov 13, 2005

### jdstokes

I don't understand this. If the energy difference ($3E_1$ say) has to precisely equal to the energy of the incident photon, what happens if the energy of the photon is arbitrarily close (but not equal) to $3E_1$? How does nature decide whether or not to absorb the photon?

5. Nov 13, 2005

### SpaceTiger

Staff Emeritus
But there are no "intermediate" energy levels in the well that correspond to an energy difference of E1. There is no process to create the new photon.

In practice, absorption lines are not infinitely thin, as this analysis might suggest. For a single absorber, the dominant form of broadening comes from the fact that the states are not truly stationary (i.e. time-independent). You get a spread in energies from the energy-time uncertainty principle:

$$\Delta E \sim \frac{h}{\Delta t}$$

This is called "Natural Broadening". Roughly speaking, any photon within that amount of the appropriate energy can be absorbed.

6. Nov 13, 2005

### Physics Monkey

If you shine monochromatic light on an infinite square well system, even if the light is "off resonance" i.e. it's frequency doesn't match any of the transistion frequencies, then the light beam can still cause transitions from the ground state to the excited states. However, these transtions are off resonant in general and proceed at a greatly reduced rate. As the light beam is brought into resonance, the transition rate will spike (limited at the very least by the fact that the light beam isn't truly monochromatic). So in the simplest approximation nothing happens when the light is off resonant but what it really means is that very little happens compared to the case of resonant absorbtion. Look up the Rabi problem (the driven two level problem) for further clarification.

In my opinion, when using the photon concept, you can't look too deeply unless you are prepared to go all the way and use the full quantized theory.

Last edited: Nov 13, 2005
7. Nov 14, 2005

### jdstokes

Thanks for the clarification SpaceTiger and Physics Monkey :).