# Finding wavelength of visible light in a well

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1. Oct 29, 2016

1. The problem statement, all variables and given/known data
Determine what colors of visible light would be absorbed by electrons in an infinite well, N = 3.1 nm. The effective mass for an electron is one-fifteenth of the standard electron mass.

2. Relevant equations

En = n2h2/(8mL2)

E = hf
f = c/λ

3. The attempt at a solution
E1 = 9.4039*10-17 J
E2 = 3.7616*10-16 J
E3 = 8.4635*10-16 J
E4 = 1.5046*10-15 J

ΔE1-2 = 70.5 nm
ΔE2-3 = 42.3 nm

Nothing in the visible.

Last edited: Oct 29, 2016
2. Oct 29, 2016

### Simon Bridge

Please show your reasoning - all you've given me is a bunch of letters and numbers.
For instance - what energies of photon can be absorbed by the electron in the ground state?

3. Oct 30, 2016

h = 6.626*10^-34
m2-kg/s
An electron can only absorb decrite (quantized) energies of a photon.
E2 − E1 = E ⇒ 70.5 nm
After using f = E/h, then λ = c/f.
Or λ = c/(E/h) = c*h/ΔE = 70.5 nm.

Since this is the highest wavelength that can be absorbed by the electron, and visible light is a higher frequency (a minimum of 400 nm), no colors are absorbed.

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4. Oct 30, 2016

### ehild

Your energy values are too high. Check the calculations.

5. Oct 30, 2016

En = n2h2/(8mL2)
En = n2(6.626e-34)2/(8(9.11e-31/15)(3.1e-9)2)
En = 9.4029*10-20n2 J

And then for the rest:
E = E2-1 = 2.82087e-19
λ = ch/E = 3e8*6.626e-34/2.82087e-19
λ = 7.0468e-7 = 704.68 nm

So red.

Last edited: Oct 30, 2016
6. Oct 30, 2016

### ehild

7. Oct 30, 2016

Is it not just taking the effective mass, me/15, as given in the question statement?

8. Oct 30, 2016

### ehild

I do not see 15 in your formula.

9. Oct 30, 2016

Sorry, it was in the Wolfram Alpha link, but I forgot to change it in the writing here. (I'm trying to get this done. I've spent all weekend on this one assignment...reading...learning...but not getting much gainful progress on the actual questions...So ya, I'm trying to go not as slow as I've been doing.)

10. Oct 30, 2016

### ehild

OK, so are there any visible wavelengths?

11. Oct 30, 2016

700ish = Dark red.

12. Oct 30, 2016

### ehild

All right. Any more?

13. Oct 30, 2016

Light? Oh, yes let me check...

14. Oct 30, 2016

### ehild

The electron can be on the second level with some probability...

15. Oct 30, 2016

ΔE2-1 = 2.82087e-19
ΔE3-2 = 4.70146e-19
ΔE4-3 = 6.58204e-19

h = m2-kg/s
6.626e-34

λ = c*h/ΔE

ΔE2-1 ⇒ λ2-1 = 705 nm (dark red)
ΔE3-2 ⇒ λ3-2 = 423 nm (purple)
ΔE4-3 ⇒ λ4-3 = 302 nm (UV ~ not visible light)

Thank you very much ehild for helping!

16. Oct 30, 2016

### ehild

You are welcome