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Hullo,

I've got a relatively simple question in my homework and yet I can't seem to find a proper answer

I need to find the confinement energy of an electron in a Deuterium nucleus with the radius of $3\cdot10^{-15} m$ and am told that the resulting energy should be around a few MeV

I first tried to solve it myself by using the Heisenberg relation and the relativistic energy formula:

$\Delta p \approx \frac{\hbar}{2\Delta x} = 1.758\cdot10^{-16} \frac{kg\cdot m}{s}$

$E = \sqrt{E_0^2+p^2c^2} = 3.29\cdot 10^{11} \text{eV}$,

which obviously doesn't fit the desired range

I also found this solution on Hyperphysics, which I don't quite get but I tried anyways and got

$\Delta p = 2.209\cdot 10^{-19}$

$E = 1.67\cdot 10^{11} \text{eV}$,

which doesn't bring me any closer to the desired answer

How should I go about solving this?

Thanks in advance

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  • $\begingroup$ Are you sure your homework is asking for confinement in the nucleus? From your link, it seems that it shouldn't even be possible. $\endgroup$
    – cxx
    Commented Nov 19, 2018 at 0:09
  • $\begingroup$ @HantingZhang yes. A literal translation: "What energy must an electron have if it were a part of the nucleus?" $\endgroup$
    – Andrii Kozytskyi
    Commented Nov 19, 2018 at 0:13

1 Answer 1

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Using just $\Delta x \Delta p = \hbar/2$, and $E = \Delta P^{2}/2m_e$, I get 1,058 MeV.

If I use the mass of a proton, I get 0.5764 MeV.

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  • $\begingroup$ Ah, I see. I believe I misunderstood the question, I think they meant that the nucleon binding energy should be around a MeV, and the electron is no nucleon. Thank you for your help! $\endgroup$
    – Andrii Kozytskyi
    Commented Nov 19, 2018 at 6:17

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