photon

[logistic_guy]

An electron trapped in an infinitely deep square well has a ground-state energy $\displaystyle E = 8.0 \ \text{eV}$. $\displaystyle \bold{(a)}$ What is the longest wave-length photon this system can emit, and $\displaystyle \bold{(b)}$ what is the width of the well?

 

[khansaheb]

An electron trapped in an infinitely deep square well has a ground-state energy $\displaystyle E = 8.0 \ \text{eV}$. $\displaystyle \bold{(a)}$ What is the longest wave-length photon this system can emit, and $\displaystyle \bold{(b)}$ what is the width of the well?

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

READ BEFORE POSTING​

Please share your work/thoughts about this problem

 

[logistic_guy]

$\displaystyle \bold{(a)}$

The photon will obtain the longest wave length when it has the least energy. In infinitely deep square well, this happens when the electron jumps from the first excited state level $\displaystyle n = 2$ to the ground state level $\displaystyle n = 1$.

We start with the photon energy formula.

$\displaystyle E = hf = \frac{hc}{\lambda} = \frac{1.24 \times 10^3 \ \text{eV} \cdot \ \text{nm}}{\lambda}$

In future Episodes, I will explain why $\displaystyle hc = 1.24 \times 10^3 \ \text{eV} \cdot \ \text{nm}$

We can use the same formula to find the change in energy between the level $\displaystyle n = 2$ and the level $\displaystyle n = 1$.

Then,

$\displaystyle \Delta E = \frac{1.24 \times 10^3 \ \text{eV} \cdot \ \text{nm}}{\lambda}$

From accumulated knowledge, we know that the energy levels formula for the electron in an infinite square well is:

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

Then,

$\displaystyle \Delta E = \left(n^2_2 - n^2_1\right)E_1 =\frac{1.24 \times 10^3 \ \text{eV} \cdot \ \text{nm}}{\lambda}$

This gives:

$\displaystyle \lambda =\frac{1.24 \times 10^3 \ \text{eV} \cdot \ \text{nm}}{\left(n^2_2 - n^2_1\right)E_1} = \frac{1.24 \times 10^3 \ \text{eV} \cdot \ \text{nm}}{\left(2^2 - 1^2\right)8 \ \text{eV}} = 51.7 \ \text{nm}$

 

[logistic_guy]

$\displaystyle \bold{(b)}$ what is the width of the well?

We can use the energy levels formula to find that.

$\displaystyle E_n = \frac{n^2h^2}{8mL^2}$

where $\displaystyle h$ is Planck's constant, $\displaystyle m$ is the rest mass of an electron, and $\displaystyle L$ is the width of the well.

Also the problem has already given us the value of $\displaystyle E_1 = 8 \ \text{eV}$.

Then,

$\displaystyle E_1 = \frac{h^2}{8mL^2}$

$\displaystyle 8 \ \text{eV} = \frac{(6.63 \times 10^{-34} \ \text{J} \cdot \text{s})^2}{8(9.11 \times 10^{-31} \ \text{kg})L^2}$

But the units don't match up, so let us get rid of the $\displaystyle \text{eV}$.

$\displaystyle 8 \ \text{eV} \times \frac{1.602 \times 10^{-19} \ \text{J}}{\text{eV}}= \frac{(6.63 \times 10^{-34} \ \text{J} \cdot \text{s})^2}{8(9.11 \times 10^{-31} \ \text{kg})L^2}$

This gives:

$\displaystyle L = 2.2 \times 10^{-10} \ \text{m} = \color{blue} 220 \ \text{pm}$