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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.

Step-by-Step

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.