What is the minimum kinetic energy of a proton in a medium-sized nucleus having a diameter of $8.0 \times 10^{-15}$ m?

Strategy We will use the uncertainty principle just as we did in Example 5.10, where we found the minimum kinetic energy for an electron in a nucleus. We use the uncertainty principle to find the uncertainty Δp and then use this value to determine the minimum kinetic energy.

Question

What is the minimum kinetic energy of a proton in a medium-sized nucleus having a diameter of [latex]8.0 imes 10^{-15}[/latex] m?

Strategy We will use the uncertainty principle just as we did in Example 5.10, where we found the minimum kinetic energy for an electron in a nucleus. We use the uncertainty principle to find the uncertainty Δp and then use this value to determine the minimum kinetic energy.

Accepted Answer

We start with the uncertainty principle involving momentum.

[latex]\begin{aligned}\Delta p \Delta x & \geq \frac{\hbar}{2} \\Delta p & \geq \frac{\hbar}{2 \Delta x}=\frac{6.58 imes 10^{-16} eV \cdot s }{2\left(8.0 imes 10^{-15} m ight)} \\Delta p & \geq 0.041 eV \cdot s / m\end{aligned}[/latex]

The momentum p must be at least as large as Δp. Hence [latex]p_{\min }=\Delta p ext {, and we have for } p_{\min } c[/latex]:

[latex]p_{\min } c=(0.041 eV \cdot s / m )\left(3.0 imes 10^{8} m / s ight)=12 MeV[/latex]

Because 12 MeV is only about 1% of the proton’s rest energy (938 MeV, see inside front cover), we can treat the problem nonrelativistically. The kinetic energy of a proton in this nucleus must be at least as large as

[latex]K=\frac{\left(p_{\min } ight)^{2}}{2 m}=\frac{\left(p_{\min } c ight)^{2}}{2 m c^{2}}=\frac{(12 MeV )^{2}}{2(938 MeV )}=0.08 MeV[/latex]

which is an entirely reasonable experimental value. The result for a neutron would be similar.

Question 12.1: What is the minimum kinetic energy of a proton in a medium-s...

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