Two nuclei of mass number 3 combine with another nucleus of mass number 4 to yield a nucleus of mass number 10

Question

Two nuclei of mass number 3 combine with another nucleus of mass number 4 to yield a nucleus of mass number 10. If the binding energy per nucleon for the mass number 3, 4 and 10 are 5.6 MeV, 7.4 MeV and 6.1 MeV, respectively, then in the process, ${ΔMc}^{2}$ = ________ MeV. [2026]

Smart Achievers · Accepted Answer

(3)

$2^{3} X + {}^{4}Y \to {}^{10}Z$

$Δ m c^{2} = B E_{prod} - B E_{React}$

$Δ m c^{2} = [2 \times 3 \times 5.6 + 4 \times 7.4 - 6.1 \times 10] = 2.2 MeV$

$| Δ m c^{2} |$ $= 2.2 MeV$

Solution

Q.
Two nuclei of mass number 3 combine with another nucleus of mass number 4 to yield a nucleus of mass number 10. If the binding energy per nucleon for the mass number 3, 4 and 10 are 5.6 MeV, 7.4 MeV and 6.1 MeV, respectively, then in the process, ${ΔMc}^{2}$ = ________ MeV. [2026]

1 6.9

2 7.9

3 2.2

4 4.3

Ans.
(3)

$2^{3} X + {}^{4}Y \to {}^{10}Z$

$Δ m c^{2} = B E_{prod} - B E_{React}$

$Δ m c^{2} = [2 \times 3 \times 5.6 + 4 \times 7.4 - 6.1 \times 10] = 2.2 MeV$

$| Δ m c^{2} |$ $= 2.2 MeV$

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