The binding energy per nucleon of ${}_{3}^{7}Li$ and ${}_{2}^{4}He$ nuclei are $5.60 ext{ MeV}$ and $7.06

Question

The binding energy per nucleon of ${}_{3}^{7}Li$ and ${}_{2}^{4}He$ nuclei are $5.60 	ext{ MeV}$ and $7.06 	ext{ MeV}$, respectively. In the nuclear reaction ${}_{3}^{7}Li + {}_{1}^{1}H ightarrow {}_{2}^{4}He + {}_{2}^{4}He + Q$, the value of energy Q released is:

Accepted Answer

1. **Concept:** The energy released ($Q$) in a nuclear reaction is equal to the difference between the total binding energy of the products and the total binding energy of the reactants .
   $$ Q = (BE)_{	ext{products}} - (BE)_{	ext{reactants}} $$
2. **Calculate Total Binding Energies:**
   - **Reactants:**
     - Lithium (${}_{3}^{7}Li$): Has 7 nucleons. $BE_{Li} = 7 	imes 5.60 	ext{ MeV} = 39.20 	ext{ MeV}$.
     - Hydrogen (${}_{1}^{1}H$): Is a single proton, so its binding energy is zero ($BE_{H} = 0$).
     - Total $BE_{	ext{reactants}} = 39.20 + 0 = 39.20 	ext{ MeV}$.
   - **Products:**
     - Helium (${}_{2}^{4}He$): Has 4 nucleons. $BE_{He} = 4 	imes 7.06 	ext{ MeV} = 28.24 	ext{ MeV}$.
     - There are two Helium nuclei. Total $BE_{	ext{products}} = 2 	imes 28.24 	ext{ MeV} = 56.48 	ext{ MeV}$.
3. **Calculate Q-value:**
   $$ Q = 56.48 	ext{ MeV} - 39.20 	ext{ MeV} $$
   $$ Q = 17.28 	ext{ MeV} \approx 17.3 	ext{ MeV} $$

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