The binding energy per nucleon of 73Li and 42He nuclei are 5.60 MeV and 7.06 MeV respectively. In the nuclear reaction 73Li + 11H 42He + 42He + Q the value of energy Q released is
- 19.6 MeV
- -2.4 MeV
- 8.4 MeV
- 17.3 MeV
Answer
(4) 17.3 MeV
Solution
Binding energy of 73Li = 7 5.60 = 39.20 MeV
Binding energy of two 42He = 2 4 7.06 = 56.48 MeV
(binding energy of 11H is zero)
Q = 56.48 - 39.20 = 17.28 MeV 17.3 MeV
Answer: (4)