For the ${ }^{40}$ Ca nucleus, we have:
- Total charge \( Q = 20e \), where \( e \) is the elementary charge (\( e \approx 1.6 \times 10^{-19} \, \text{C} \)).
- The radius \( R \) of the nucleus can be estimated using the formula \( R = R_0 A^{1/3} \), where \(
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