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Compare the average binding energy of a nucleon in ${^{23}_{11}\textrm{Na}}$ to that in ${^{24}_{11}\textrm{Na}}$, using Appendix B.

Source: Giancoli, Douglas C., Physics: Principles with Applications, 7th Edition, 2014.

Quick Answer: 

${}^{23}Na: 8.11 \textrm{ MeV/nucleon, }$ ${}^{24}Na: 8.06 \textrm{ MeV/nucleon}$

Giancoli 7th Edition, Chapter 30, Problem 19


Chapter 30, Problem 19 is solved.

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