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The Binding energy per nucleon of 3 7 ^7_3 3 7 Li and 2 4 ^4_2 2 4 He are 5.60 5.60 5.60 MeV and 7.06 7.06 7.06 MeV, respectively. In the nuclear reaction \newline 3 7 ^7_3 3 7 Li + 1 1 + ^1_1 + 1 1 H → 2 4 \to ^4_2 → 2 4 He + 2 4 + ^4_2 + 2 4 He + Q , + Q, + Q , the value of energy Q released is
Explanation To solve this problem, we need to calculate the energy released in the given nuclear reaction. \\
The reaction is: 3 7 \\
^7_3 3 7 Li + 1 1 + ^1_1 + 1 1
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H
He
He
First, we calculate the total binding energy of the reactants and products.
Binding energy per nucleon of
Li
is
MeV.
Number of nucleons in
Li
is
Total binding energy of
Li
= 5.60 × 7 = 39.2 = 5.60 \times 7 = 39.2 = 5.60 × 7 = 39.2 MeV.
Binding energy per nucleon of
H
is
MeV (since it's a single proton).
Total binding energy of
H
MeV.
Total binding energy of reactants:
39.2 + 0 = 39.2 \\
39.2 + 0 = 39.2 39.2 + 0 = 39.2 MeV.
Binding energy per nucleon of
He
is
MeV.
Number of nucleons in each
He
is
Total binding energy of one
He
= 7.06 × 4 = 28.24 = 7.06 \times 4 = 28.24 = 7.06 × 4 = 28.24 MeV.
Since there are two
He
nuclei produced, the total binding energy of products is:
2 × 28.24 = 56.48 \\
2 \times 28.24 = 56.48 2 × 28.24 = 56.48 MeV.
Total binding energy of products:
MeV.
The energy released
is the difference between the total binding energy of the products and the reactants:
Q = 56.48 − 39.2 = 17.28 \\
Q = 56.48 - 39.2 = 17.28 Q = 56.48 − 39.2 = 17.28 MeV.
Therefore, the value of energy
released is approximately
MeV.
This corresponds to Option 4.