Question

# Estimate the lowest possible energy of a neutron contained in a typical nucleus of radius $1.0 \times 10^{-15} \mathrm{m}$ [Hint: a particle can have an energy at least as large as its uncertainty.]

Solution

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If we consider the radius of the atom to correspond to the uncertainty in its position, we can find the uncertainty in momentum.

$\Delta p = \frac{\hbar}{\Delta x} = \frac{1.055 \times 10^{-34} \text{ m^2 kg/s}}{10^{-15} \text{ m}} = 1.055 \times 10^{-19} \text{ m kg/s}$

Using the hint from the task, we can find the minimum energy using the uncertainty in momentum.

$E = \frac{(\Delta p)^2}{2m} = \frac{(1.055 \times 10^{-19} \text{ m kg/s})^2}{2 \cdot 1.67 \times 10^{-27} \text{ kg}} = 3.33 \times 10^{-12} \text{ J} = \boxed{ 20.8 \text{ MeV} }$

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