Nucleus: Practice Problem & Solution
The binding energy per nucleon of $^7_3Li$ and $^4_2He$ nuclei are 5.60 MeV and 7.06 MeV, respectively. In the nuclear reaction $^7_3Li + ^1_1H \rightarrow ^4_2He + ^4_2He + Q$ the value of energy Q released is: (2014)
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:
Total binding energy of reactants = $(7 \times 5.60) + 0 = 39.20$ MeV. Total binding energy of products = $2 \times (4 \times 7.06) = 56.48$ MeV. Energy released $Q = BE_{products} - BE_{reactants} = 56.48 - 39.20 = 17.28$ MeV $\approx 17.3$ MeV.
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