Finding wavelength of visible light in a well

[adamaero]

Homework Statement

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.

Homework Equations

E_n = n²h²/(8mL²)

E = hf
f = c/λ

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

ΔE_1-2 = 70.5 nm
ΔE_2-3 = 42.3 nm

Nothing in the visible.

 

[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?

 

[adamaero]

h = 6.626*10^-34
m²-kg/s

Please show your reasoning.

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.

 

[ehild]

Homework Equations

E_n = n²h²/(8mL²)

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

Your energy values are too high. Check the calculations.

 

[adamaero]

Your energy values are too high. Check the calculations.

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 = E_2-1 = 2.82087e-19
λ = ch/E = 3e8*6.626e-34/2.82087e-19
λ = 7.0468e-7 = 704.68 nm

So red.

 

[ehild]

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

Is the mass correct?

 

[adamaero]

Is the mass correct?

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

 

[ehild]

Am I not just taking the effective mass, m_e/15, as given in the question statement?

I do not see 15 in your formula.

 

[adamaero]

I do not see it in your formula.

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.)

 

[ehild]

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...)

OK, so are there any visible wavelengths?

 

[adamaero]

OK, so are there any visible wavelengths?

700ish = Dark red.

 

[ehild]

All right. Any more?

 

[adamaero]

All right. Any more?

Light? Oh, yes let me check...

 

[ehild]

Light? Oh, yes let me check...

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

 

[adamaero]

ΔE_2-1 = 2.82087e-19
ΔE_3-2 = 4.70146e-19
ΔE_4-3 = 6.58204e-19

h = m2-kg/s
6.626e-34

λ = c*h/ΔE

ΔE_2-1 ⇒ λ_2-1 = 705 nm (dark red)
ΔE_3-2 ⇒ λ_3-2 = 423 nm (purple)
ΔE_4-3 ⇒ λ_4-3 = 302 nm (UV ~ not visible light)

Thank you very much ehild for helping!

 

[ehild]

You are welcome