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A problem in my textbook states:

Using the uncertainty principle, find the minimum value (in MeV) of the kinetic energy of a nucleon confined within a nucleus of radius $R = 5 \times 10^{-15}$m.

Given the Heisenberg uncertainty principle: $$ \Delta x \Delta p \geqslant \frac{\hbar}{2} $$

I understand one can, using either classical or relativistic formulae for kinetic energy, and plugging in the value for $R$, solve for the energy and find a solution of the form: $$ \Delta pc \geqslant \frac{\hbar c}{2\Delta x} $$ or $$ \frac{\Delta p^2}{2m} \geqslant \frac{\hbar^2}{2m\Delta x^2} $$

Now I fail to understand how these solutions (given to be correct by the textbook) actually imply a minimum energy. To me it seems that these solutions rather imply a minimum uncertainty in energy. Does this minimum energy-uncertainty imply an equal minimum energy, and if so, why?

EDIT:

I read elsewhere that since it is implied that the uncertainty of a quantity (in this case energy) is smaller than its magnitude, the minimum uncertainty also serves as a lower bound for the magnitude. I should therefore rephrase my question as follows:

Is it usually the case that one can assume such a quantity to have a larger magnitude than uncertainty, and if not, can this be assumed for a general case such as this one?

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