Discussion
The mean momentum \(\bar{p}\) of a nucleon in a nucleus of mass number A and atomic number Z depends on A, Z as
Answer: C
The key gives \(\bar{p} \propto A^{-1/3}\).
The key follows the uncertainty-principle estimate. A nucleon is confined within the nucleus, whose radius is \(R = R_{0}A^{1/3}\), so its momentum spread is of order
\[ \begin{aligned} \bar{p} &\sim \frac{\hbar}{R} = \frac{\hbar}{R_{0}A^{1/3}} \\ &\propto A^{-1/3} \end{aligned} \]
This is the textbook estimate the question intends; a fuller Fermi-gas treatment, which accounts for all A nucleons filling momentum states, gives a typical momentum that is nearly independent of A. Among the options, \(A^{1/3}\) has the dependence inverted, and the forms in Z or \(AZ\) have no basis, since the size of the nucleus depends on the total number of nucleons.
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