Nuclear Physics And Radioactivity Question 220
Question: The nuclear reaction $ ^{2}H{{+}^{2}}H\to {{,}^{4}}He $ (mass of deuteron = 2.0141 a.m.u. and mass of He = 4.0024 a.m.u.) is [Orissa JEE 2002]
Options:
A) Fusion reaction releasing 24 MeV energy
B) Fusion reaction absorbing 24 MeV energy
C) Fission reaction releasing 0.0258 MeV energy
D) Fission reaction absorbing 0.0258 MeV energy
Show Answer
Answer:
Correct Answer: A
Solution:
Total mass of reactants $ =(2.0141)\times 2=4.0282\ amu $ Total mass of products $ =4.0024\ amu $ Mass defect $ =4.0282\ amu-4.0024\ amu $ $ =0.0258\ amu $
$ \therefore $ Energy released $ E=931\times 0.0258=24\ MeV $
BETA
