Nucleus: Practice Problem & Solution
The mass of a $ _3^7Li $ nucleus is 0.042 u less than the sum of the masses of all its nucleons. The binding energy per nucleon of $ _3^7Li $ nucleus is nearly: (2010 Pre)
Solution Explained:
To solve this problem, we apply the core principles of Nucleus. Understanding the underlying formula is key to arriving at the correct answer below:
Mass defect $ \Delta m = 0.042 $ u. Total binding energy $ = \Delta m \times 931.5 $ MeV = $ 0.042 \times 931.5 = 39.123 $ MeV. Binding energy per nucleon = $ 39.123 / 7 \approx 5.6 $ MeV.
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