Question

The binding energy per nucleon of \(Li\) and \(He\) nuclei are \(5.60\, MeV\) and \(7.06\, MeV\) respectively. In the nuclear reaction \(^7_3Li+_1^1H \rightarrow  _2^4He + _2^4He+Q\) the value of energy Q released is

• A. 19.6 MeV
• B. -2.4 MeV
• C. 8.4 MeV
• D. 17.3 MeV

Answer 1

The Correct Option is D

To solve the problem of finding the energy \( Q \) released in the nuclear reaction \( ^7_3\text{Li} + \, ^1_1\text{H} \rightarrow \, ^4_2\text{He} + \, ^4_2\text{He} + Q \), we'll use the concept of binding energy per nucleon.

The binding energy per nucleon for:

• \(^7_3\text{Li} = 5.60\, \text{MeV}\)
• \(^4_2\text{He} = 7.06\, \text{MeV}\)

Step-by-Step Calculation:

1. The total binding energy of \( ^7_3\text{Li} \):
   • The number of nucleons in \( \text{Li} = 7 \).
   • Total binding energy = Binding energy per nucleon \(\times\) Number of nucleons = \( 5.60 \times 7 = 39.2\, \text{MeV} \).
2. The total binding energy of \( ^1_1\text{H} \) is negligible, typically around \( 0 \) MeV, because it is a single proton and does not contribute significantly to binding energy.
3. The total binding energy of the products has two \( ^4_2\text{He} \) nuclei:
   • The number of nucleons in \( \text{He} = 4 \).
   • Binding energy per nucleus = Binding energy per nucleon \(\times\) Number of nucleons = \( 7.06 \times 4 = 28.24\, \text{MeV} \) per \( \text{He} \) nucleus.
   • Since there are two \( ^4_2\text{He} \) nuclei, total = \( 28.24 + 28.24 = 56.48\, \text{MeV} \).
4. Energy \( Q \) released is the difference in binding energies of products and reactants:
   • \( Q = \text{Total binding energy of products} - \text{Total binding energy of reactants} \)
   • \( Q = 56.48\, \text{MeV} - 39.2\, \text{MeV} = 17.28\, \text{MeV} \approx 17.3\, \text{MeV} \).

Hence, the energy \( Q \) released in the reaction is approximately \( 17.3\, \text{MeV} \), which confirms the correct answer is 17.3 MeV.