Nuclear Physics And Radioactivity Question 217
Question: The binding energy per nucleon of deuterium and helium atom is 1.1 MeV and 7.0 MeV. If two deuterium nuclei fuse to form helium atom, the energy released is [Pb. PMT 2001; CPMT 2001; AIEEE 2004]
Options:
A) 19.2 MeV
B) 23.6 MeV
C) 26.9 MeV
D) 13.9 MeV
Show Answer
Answer:
Correct Answer: B
Solution:
$ _{1}H^{2}+ _{1}H^{2}\to _{2}He^{4}+ $ energy
Binding energy of a $ _{1}H^{2} $ deuterium nuclei $ =2\times 1.1=2.2\ MeV $ Total binding energy of two deuterium nuclei $ =2.2\times 2=4.4\ MeV $ Binding energy of a $ _{2}He^{4} $ nuclei $ =4\times 7=28\ MeV $ So, energy released in fusion $ =28-4.4=23.6\ MeV $
BETA
