QUANTUM PHYSICS FUNDAMENTALS - Chapter 6, Exercise 1 Solution
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Minimum Momentum Uncertainty for a Proton in a Nucleus

PROBLEM
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A proton is confined to a region with position uncertainty
Delta x = 2 x 10^-15 m (roughly the size of an atomic nucleus).
Calculate the minimum possible uncertainty in its momentum.

SOLUTION
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Using the real, rigorous uncertainty relation:

    Delta p >= hbar / (2 Delta x)

Substituting hbar = 1.0546 x 10^-34 J.s and Delta x = 2 x 10^-15 m:

    Delta p >= (1.0546 x 10^-34) / (2 x 2 x 10^-15)
    Delta p >= (1.0546 x 10^-34) / (4 x 10^-15)
    Delta p >= 2.6365 x 10^-20 kg.m/s

ANSWER: The minimum possible uncertainty in the proton's momentum is
approximately 2.64 x 10^-20 kg.m/s.

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WHY THIS WORKS AS AN ANSWER

This result is roughly 100,000 times larger than the chapter's own
electron-in-an-atom example (5.27 x 10^-25 kg.m/s), even though the
proton's own confinement region (2 x 10^-15 m, a nucleus) is much
SMALLER than the electron's (1 x 10^-10 m, an atom). This is a real,
direct illustration of the inverse relationship in the uncertainty
principle: confining a particle to a smaller region of space
(smaller Delta x) necessarily forces a genuinely larger minimum
uncertainty in its momentum (larger Delta p), regardless of the
particle's own mass. This same real relationship is part of why
protons and neutrons confined within an atomic nucleus carry
substantial real kinetic energy even at a temperature of absolute
zero - a genuine, physical consequence of the uncertainty principle,
not a measurement limitation.