# Homework Help: Where did pi^2 go in this QM j/eV unit conversion?

1. Jun 23, 2011

### timn

1. The problem statement, all variables and given/known data

Consider an electron of mass m=9*10^-30 kg in an infinite box of dimension a=10^-9 m. What is the energy difference between the ground state and the first excited state? Express your answer in eV.

2. Relevant equations

$$E_n = \frac{\hbar^2\pi^2n^2}{2ma^2}$$

3. The attempt at a solution

$$E_2 - E_1 = 3E_1 = 3 \frac{\hbar^2\pi^2}{2ma^2} \approx \frac{ (1.054\times10^{-34} Js )^2 \pi^2 }{ 2(9\times 10^{-30} kg)(10^{-9} m)^2 } (6.24\times 10^{18} J/eV) \approx 1.14 eV$$

The answer is supposed to be almost exactly 1/pi^2 of mine: 0.115 eV. Why? Am I being confused by a unit conversion?

Edit: I'm having a similar problem with the follow-up question: "Suppose the transition from the state n=2 to the state n=1 is accompanied by the emission of a photon, as given by the Bohr rule. What is its wavelength?"

$$\Delta E=h\frac{c}{\lambda} \Leftrightarrow \lambda = \frac{hc}{\Delta E} = \frac{2 \pi \hbar c}{\Delta E} \approx \frac{2\pi ( 6.6\times 10^{-16} eV s )( 3\times 10^8 m/s ) }{\Delta E} = \frac{2\pi ( 1.98 \times 10^{-7} eV m ) }{\Delta E}$$

According to the key, the numerator in the latter is supposed to be 2pi*2.6*10^-7 eV m. What causes this factor?

$$\frac{2.6}{1.98}\approx\pi$$

Last edited: Jun 23, 2011
2. Jun 23, 2011

### Clever-Name

First of all why is the mass of the electron $9*10^{-30}$... It's $~9.11*10^{-31}$ I'm going to assume you're using E-30 just as a rough approximation but that actually shocks me. I will use what you wrote in my calculations

You can simplify the energy formula to $$E = \frac{3h^{2}}{8ma^{2}}$$
Recall ${\hbar} = \frac{h}{2{\pi}}$
First to make things a little simpler. I noticed you didn't carry the 3 through in your post but I assume you used it in your calculation in the end.

Calculating that out I get:
$$E = \frac{3(6.626*10^{-34})^{2}}{8(9*10^{-30})(10^{-9})^{2}}$$
$$E = 1.829 * 10^{-20} J$$

Converting to eV I get:

$$E = 0.114eV$$

Is it possible you were using the value for h and not hbar when doing your calculation?

3. Jun 23, 2011

### vela

Staff Emeritus

You can avoid some of the nuisance and confusion of unit conversions by sticking with units of electron-volts and nanometers and using the combinations
\begin{align*}
hc &= 1240~\mathrm{nm~eV} \\
\hbar c &= 197~\mathrm{nm~eV} \\
m_ec^2 &= 511~\mathrm{keV}
\end{align*}
where $m_e$ is the mass of the electron. These combinations occur often enough that it's worth memorizing them.

You may need to slightly rewrite some expressions, though, by introducing factors of c. For example, for the energy levels, you'd do
$$E_n = \frac{\hbar^2 \pi^2 n^2}{2ma^2} = \frac{(\hbar c)^2 \pi^2 n^2}{2(mc^2)a^2} = \frac{(197~\mathrm{nm~eV})^2\pi^2 n^2}{2(511~\mathrm{keV})a^2}$$
You can see if a is in nanometers, the units conveniently reduce down to eV, which is what you wanted.

4. Jun 23, 2011

### vela

Staff Emeritus
Oh, I didn't notice the electron mass in the problem isn't the real electron mass. If it's not a simple typo, it's off by a factor of 10, which is why your answer and the key's answers are off by the same factor. Despite what you showed in your work, you apparently used the real electron mass in your calculation. It has nothing to do with factors of pi. It's just a coincidence that $\pi^2$ is almost equal to 10.

5. Jun 23, 2011

### timn

Sorry about missing the 3 and the electron mass -- it's supposed to say 0.9*10^-30. I'll look through your calculations and simplifications soon. Thank you very much for the input!

6. Jun 23, 2011

### timn

Thanks for your suggestions on the unit rearrangements, vela. I'll keep that in mind and see if it makes things easier.

The answers were found in a shady .doc file from scribd, and this would not be the first error. Thanks!