First slide

Nuclear Binding energy

Question

The binding energy per nucleon of ${}_{3}^{7}L i$ and ${}_{2}^{4}H e$ Nuclei are 5.60 MeV and 7.06 MeV respectively. In the nuclear reaction
${}_{3}^{7}{Li} + {}_{1}^{1}H \to {}_{2}^{4}{He} + {}_{2}^{4}{He} + Q$
the value of energy Q released is

Easy

A

19.6 MeV
B

-2.4 MeV
C

8.4 MeV
D

17.3 MeV

Solution

$Binding energy of {}_{3}^{7}{Li} nucleus = 7 \times 5.60 MeV = 39.2 MeV$
$Binding energy of {}_{2}^{4}{Henucleus} = 4 \times 7.06 MeV = 28.24 MeV$
the reaction is
${}_{3}^{7}{Li} + {}_{1}^{1}H \to 2 ({}_{2}^{4}{He}) + Q$
$∴ Q = 2 (BE of {}_{2}^{4}{He}) - (BE of {}_{3}^{7}{Li})$
$= 2 \times 28.24 MeV - 39.2 MeV = 56.48 MeV - 39.2 MeV$
$= 17.28 MeV \approx 17.3 MeV$

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