Atoms And Nuclei Question 159
Question: The binding energy of deuteron $ _{1}^{2}H $ is 1.112 MeV per nucleon and an $ \alpha - $ particle $ _{2}^{4}He $ has a binding energy of 7.047 MeV per nucleon. Then in the fusion reaction $ _{1}^{2}H+ _{1}^{2}H\to _{2}^{4}He+Q $ , the energy Q released is [MP PMT 1992; Roorkee 1994; IIT 1996; AIIMS 1997]
Options:
A) 1 MeV
B) 11.9 MeV
C) 23.8 MeV
D) 931 MeV
Show Answer
Answer:
Correct Answer: C
Solution:
Mass of $ _{1}H^{2}=2.01478\ a.m.u. $
Mass of $ _{2}He^{4}=4.00388\ a.m.u. $
Mass of two deuterium $ =2\times 2.01478=4.02956 $
Energy equivalent to $ 2_{1}H^{2} $
$ =4.02956\times 1.112\ MeV=4.48\ MeV $
Energy equivalent to $ _{2}H^{4} $
$ =4.00388\times 7.047\ MeV=28.21\ MeV $
Energy released $ =28.21-4.48=23.73\ MeV $ = 24 MeV
