(adsbygoogle = window.adsbygoogle || []).push({}); 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

[tex] E_n = \frac{\hbar^2\pi^2n^2}{2ma^2} [/tex]

3. The attempt at a solution

[tex] 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

[/tex]

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?"

[tex]

\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}

[/tex]

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

[tex]\frac{2.6}{1.98}\approx\pi[/tex]

**Physics Forums | Science Articles, Homework Help, Discussion**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

**Physics Forums | Science Articles, Homework Help, Discussion**