Quantum Physics (Electron in infinite deep quantum well)

[jisbon]

Homework Statement

An electron is trapped in the ground state in an infinitely deep quantum well With
a width of 2 nm. After absorbing a photon, the electron transits to an excited state
with a de Broglie wavelength of 1 nm.
(i) Find the quantum number of the excited state.
(ii) Sketch the wave function of the excited state.
(iii) Determine the wavelength of the incident photon.

Relevant Equations

E= h^2n^2/8mL
Wavelength = 2L/n

Hi all, below is my attempt. Pretty new at quantum so do correct me if I'm wrong. Thanks

i) Since 1nm is the wavelength of the electron, Wavelength = 1nm, 1nm=2(2nm)/n , n=4?

ii) -

iii) From n=1 to n=4,
Change in E = 15h^2/8mL where m is the mass of electron and L is 2nm?
From the change in E I can find the wavelength using E(change) = 1240/nm ?

Cheers

 

[PeroK]

Relevant Equations:: E= h^2n^2/8mL

Change in E = 15h^2/8mL

Are you sure about that equation?

 

[jisbon]

Are you sure about that equation?

Since n=4 in the final state and n=1 in its initial state, wouldn't the change be 16h^2/8mL - h^2/8mL = 15h^2/8mL?

 

[PeroK]

Since n=4 in the final state and n=1 in its initial state, wouldn't the change be 16h^2/8mL - h^2/8mL = 15h^2/8mL?

The $\frac {15} 8$ is correct, but the equation is wrong otherwise.

 

[jisbon]

So I'm using the wrong equation to solve part b?
Is part a correct though?

 

[PeroK]

So I'm using the wrong equation to solve part b?
Is part a correct though?

The energy of a well is:
$$\frac{n^2h^2}{8mL^2}$$

 

[jisbon]

Oh.. Missed out the L.. So..
$\frac{15h^2}{8(9.1*10^{-31})(2*10^{-9})^2} = \frac{1240}{wavelength (nm)}$?

 

[PeroK]

Oh.. Missed out the L.. So..
$\frac{15h^2}{8(9.1*10^{-31})(2*10^{-9})^2} = \frac{1240}{wavelength (nm)}$?

I would write:
$$\lambda = \frac{8mcL^2}{15h}$$
And then put that in a calculator.