# Math Help - Units (Physics)

1. ## Units (Physics)

I'm given $E_n = \frac{\hbar^2n^2\pi^2}{2ma}$ and I had to calculate $E_1$ which the book gets

$E_1 = 6\cdot 10^{-18}J = 37~eV$

No idea how... the orig. prob says that an electron has walls of width $10^{-8}~ cm$ which I guess is our a. Then, we're given:

$\hbar = 6.58217\cdot 10^{-16}~ eV-s$

$m = 0.511~ MeV/c^2$

Guess c is light, which is: $2.99792\cdot 10^{10}~ cm/s$.

So how exactly do they get that?

2. Originally Posted by Ideasman
I'm given $E_n = \frac{\hbar^2n^2\pi^2}{2ma}$ and I had to calculate $E_1$ which the book gets

$E_1 = 6\cdot 10^{-18}J = 37~eV$

No idea how... the orig. prob says that an electron has walls of width $10^{-8}~ cm$ which I guess is our a. Then, we're given:

$\hbar = 6.58217\cdot 10^{-16}~ eV-s$

$m = 0.511~ MeV/c^2$

Guess c is light, which is: $2.99792\cdot 10^{10}~ cm/s$.

So how exactly do they get that?
For starters your formula is wrong:
$E = \frac{\hbar ^2 n^2 \pi ^2}{2ma^2}$

I'd convert everything into MKS and then convert to eV at the end.

$E_1 = \frac{(1.855 \times 10^{-34}~Js)^2(1)^2 \pi ^2}{2 \cdot 9.11 \times 10^{-31}~kg \cdot (10^{-10}~m)^2}$

I get the numbers the question was talking about using this formula.

-Dan