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Chapter 12

Q. 12.1

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.


Verified Solution

We start with the uncertainty principle involving momentum.


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


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


p_{\min } c=(0.041 eV \cdot s / m )\left(3.0 \times 10^{8} m / s \right)=12 MeV


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


K=\frac{\left(p_{\min }\right)^{2}}{2 m}=\frac{\left(p_{\min } c\right)^{2}}{2 m c^{2}}=\frac{(12 MeV )^{2}}{2(938 MeV )}=0.08 MeV


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