Question

# The binding energy per nucleon of $$^7_3 Li$$ and $$^4_2 He$$ nuclei are 5.60 MeV and 7.06 Me V, respectively. In the nuclear reaction $$^7_3 Li + ^1_1 He \longrightarrow ^4_2 H + ^4_2 He + Q$$ the value of energy Q released is

A
-2.4 MeV
B
8.4 MeV
C
17.3 MeV
D
19.6 MeV

Solution

## The correct option is C 17.3 MeVThe binding energy for $$_1H^1$$ is around zero and also not give in the question so we can ignore it,$$Q=2(4\times 7.06)-7\times (5.60)$$$$=(56.48-39.2)MeV$$$$=17.28\ MeV$$$$=17.3\ MeV$$Physics

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