# Electron in one dimensional infinite square well

1. Apr 27, 2013

### whynot314

An electron in the ground state of a one-dimensional infinite square well of width 1.10 nm is
illuminated with light of wavelength 600 nm. Into which quantum state is the electron excited?

ok so I first calculated the engery of the electron in the first ground state of the square well using
$\frac{1240^{2}}{8*511000*1.10^{2}}$
I get .310 eV

Then i calculate the energy of the photon
$\frac{1240}{600}$

and i get 2.066 eV

Then i do
2.066=$\frac{1240^{2}}{8*511000*1.10^{2}}$*$n^{2}$-.310
and solve for n i get the 3rd state. that answer says it should be in the same state n=1, I am not sure what I am doing wrong.

2. Apr 27, 2013

### Staff: Mentor

If that fraction is 0.31 eV, how did you get n=3 as result? The solution differs significantly from 3 - indicating that the electron does not absorb this wavelength at all.

3. Apr 27, 2013

### whynot314

The .310 I calculated using the energy for a particle in a box E=$\frac{h^{2} c^{2}}{8*m*c^{2}*L^{2} }$*$n^{2}$, for n=1 for a the electron in ground state. I got n=3 by using E(photon)= (energy in a box) $\frac{h^{2} c^{2}}{8*m*c^{2}*L^{2} }$*$n^{2}$-Energy of the electron in ground state (.310). I just then solved for n.

so how does the ratio compare?

4. Apr 27, 2013

### whynot314

Does the photon need to have the exact energy level to excite the electron into a different state? since the ground state is .311 eV and the 1 st excited state is 1.244 eV and the 2nd excited state is 2.799 for this length of box. And because a single photon of the light has 2.066 eV and does not match any of the other excited states it does not get absorbed?

5. Apr 28, 2013

Pretty much.
Exactly.