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Thread: Units (Physics)

  1. #1
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    Units (Physics)

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

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

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

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

    $\displaystyle m = 0.511~ MeV/c^2$

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

    So how exactly do they get that?
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  2. #2
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    Quote Originally Posted by Ideasman View Post
    I'm given $\displaystyle E_n = \frac{\hbar^2n^2\pi^2}{2ma}$ and I had to calculate $\displaystyle E_1$ which the book gets

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

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

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

    $\displaystyle m = 0.511~ MeV/c^2$

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

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

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

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

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

    -Dan
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