Today I came across a question , where I was asked to find the Strength of nuclear force from Heisenberg's Uncertainty Principle...

Please can anyone help me out... I have come across a solution which I think is not correct as far as the calculations are involved...enter image description here

  • $\begingroup$ What if I want to get the magnitude of the energy associated $\endgroup$ – Abhishek Ghosh Dec 4 '18 at 16:37
  • $\begingroup$ Isn't it already written ? I didnt understand what you are asking for ? $\endgroup$ – Reign Dec 4 '18 at 16:51
  • $\begingroup$ The value comes out as 52keV and not 10MeV $\endgroup$ – Abhishek Ghosh Dec 4 '18 at 16:59
  • $\begingroup$ @ArthurMorgan $E_k$ isn't even close to 10 MeV. The calculation is ~50 keV. $\endgroup$ – Bill N Dec 4 '18 at 17:00
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    $\begingroup$ Where did you find that example? $\endgroup$ – Bill N Dec 4 '18 at 17:06

Let's try this without using macroscopic units --- sometimes even experts can do silly things with scientific notation. We'll set

  • $\hbar c \approx 0.2\rm\,GeV\,fm$ (reference)
  • $M_\text{nucleon}c^2 \approx 1\rm\,GeV$
  • $\Delta x \approx 10\rm\,fm $

Note that this is kind of a large nucleus. The effective radius of a nucleon is closer to $\rm1\,fm$, and since nucleons in a nucleus are close-packed then the radius of a nucleus with mass number $A$ is pretty close to $A^{1/3}\rm\,fm$. But since your question is about orders of magnitude, we'll leave it there.

The uncertainty in the momentum --- and therefore an order-of-magnitude estimate for the typical magnitude of the momentum --- is

\begin{align} \Delta p \Delta x &\gtrsim \hbar \\ \Delta p c &\gtrsim \frac{\hbar c}{\Delta x} = \frac{200\rm\,MeV\,fm}{10\rm\,fm} = 20\rm\,MeV \end{align}

This is small compared to the nucleon mass, so we can use the non-relativistic kinetic energy:

\begin{align} E = \frac{p^2}{2M} &= \frac{(pc)^2}{2 Mc^2} \\ &= \frac{(20\rm\,MeV)^2}{2000\rm\,MeV} = 0.2\rm\,MeV. \end{align}

Doubling $\Delta x$ brings this energy estimate down to $50\rm\,keV$, as you have correctly computed; your photographed text is wrong.

Note that the statement of the uncertainty principle,

$$ \sigma_p \sigma_x \geq \hbar / 2 $$

is an inequality, so a simple way to interpret this discrepancy is to claim that a nucleon rattling around in a nucleus isn't in a minimum-uncertainty state, which isn't too surprising. An interesting reversal of this homework problem would be to build a smarter model of a nucleon in a nucleus --- say a three-dimensional square well with the correct radius and an empirically-observed nucleon separation energy of 8-ish MeV --- and compute $\Delta x$ and $\Delta p$ for that solution, to see by what margin the uncertainty principle is actually obeyed. What you'll find is that the ground state for a square-well wavefunction is pretty close to a minimum-uncertainty wavepacket, but because of the way that the square well wavefunctions are quantized, there's no simple connection between the depth of the well $V$ and the kinetic energy $E-V \approx p^2/2m$.

For what it's worth, this approach works much better for electronic energy levels:

\begin{align} \Delta x_\text{atom} &\sim 0.1\rm\,nm \\ \Delta p c \geq \frac{\hbar c}{2\Delta x} &= \frac{197\rm\,eV\,nm}{2\times 0.1\rm\,nm} \approx 10^3\rm\,eV \\ E \sim \frac{(\Delta p c)^2}{2m_\text{electron}c^2} &= \frac{10^6\rm\,eV^2}{2\times\frac12\times10^6\rm\,eV} \approx 1\rm\,eV \end{align}

That's probably because the virial theorem guarantees that the kinetic and potential energies in the electrical interaction between an electron and its nucleus have the same magnitude, while the relationship for the nuclear potential is a lot messier.


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