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  1. #1
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    Question on potential calculation

    Hi everyone
    need help for this question, im not sure what formula to use.really appreciate if someone can help me.

    A proton with an energy of 1.5 eV impinging on a potential barrier with
    V= 15 eV and a width of 2Ň . Assume that the proton mass is equal to the
    rest mass of electron. Calculate the probability of the proton tunnelling
    through the potential barrier


    thank you in advance for all your help & support,really appreciate.
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  2. #2
    Member mathemagister's Avatar
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    Quote Originally Posted by anderson View Post
    Hi everyone
    need help for this question, im not sure what formula to use.really appreciate if someone can help me.

    A proton with an energy of 1.5 eV impinging on a potential barrier with
    V= 15 eV and a width of 2Ň . Assume that the proton mass is equal to the
    rest mass of electron. Calculate the probability of the proton tunnelling
    through the potential barrier


    thank you in advance for all your help & support,really appreciate.
    [This isn't only Math, well it's the Math of Quantum Physics]


    The probability of tunnelling is given (approximately) by P \approx e^{-2\alpha L}, where \alpha = \frac{\sqrt{2m(U_0-E)}}{h(2\pi)^{-1}}.

    The key is to multiply the whole fraction by \frac{c}{c}, (inside the numerator's square root, you should multiply by c^2). The reason for this is that, now, you get hc (a known constant) in the denominator and you get mc^2 in the numerator, which is just energy [E=mc^2].

    \alpha = \frac{\sqrt{2m(U_0-E)}}{h(2\pi)^{-1}}=\frac{\sqrt{2mc^2(U_0-E)}}{hc(2\pi)^{-1}}.

    Now simply substitute everything in, note that the constant hc is roughly 197eVnm.

    The rest mass (m) is approximately 5\cdot 10^5 eVc^{-2}

    Multiplying both sides by c^2, mc^2, which is energy, just becomes 5*10^5 eV.

    Substituting everything, alpha becomes:

    \alpha = \frac{\sqrt{2(5\cdot10^5eV)(13.5eV)}}{197eVnm} \approx 18.65 nm^{-1}

    The barrier width is 2 angstrom which is 0.2nm:

    P \approx e^{-2\alpha L} \approx e^{-2(18.65nm^{-1})(0.2nm)}

    P \approx e^{-7.46}

    And, there you go

    Hope I helped you, anderson.
    Last edited by mathemagister; February 16th 2010 at 04:35 AM.
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  3. #3
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    Dear mathemagister

    thank you so much, relly helped me.

    thank you again.
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  4. #4
    Member mathemagister's Avatar
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    Quote Originally Posted by anderson View Post
    Dear mathemagister

    thank you so much, relly helped me.

    thank you again.
    No problem. Glad to help!
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