Solved: PHYSICS Question for NEET

NEET PHYSICSEasy

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

19.6 MeV

-2.4 MeV

8.4 MeV

17.3 MeV

Step-by-Step Solution

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 . $Q = (BE)_{\text{products}} - (BE)_{\text{reactants}}$
Calculate Total Binding Energies:

Reactants:
Lithium ( ${}_{3}^{7}Li$ ): Has 7 nucleons. $BE_{Li} = 7 \times 5.60 \text{ MeV} = 39.20 \text{ MeV}$ .
Hydrogen ( ${}_{1}^{1}H$ ): Is a single proton, so its binding energy is zero ( $BE_{H} = 0$ ).
Total $BE_{\text{reactants}} = 39.20 + 0 = 39.20 \text{ MeV}$ .
Products:
Helium ( ${}_{2}^{4}He$ ): Has 4 nucleons. $BE_{He} = 4 \times 7.06 \text{ MeV} = 28.24 \text{ MeV}$ .
There are two Helium nuclei. Total $BE_{\text{products}} = 2 \times 28.24 \text{ MeV} = 56.48 \text{ MeV}$ .

Calculate Q-value: $Q = 56.48 \text{ MeV} - 39.20 \text{ MeV}$ $Q = 17.28 \text{ MeV} \approx 17.3 \text{ MeV}$

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