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

## Homework Statement

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.

## Homework Equations

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

## 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?

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

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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?

vela
Staff Emeritus
Homework Helper

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.

vela
Staff Emeritus
Homework Helper
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.