JEE Main On 8 April 2017 Question 28
Question: Two deuterons udnergo nuclear fusion to form a Helium nucleus. Energy released in this process is : [JEE Online 08-04-2017] (given binding energy per nucleon for deuteron = 1.1 MeV and for helium = 7.0 MeV)
Options:
A) 23.6 MeV
B) 25.8 MeV
C) 30.2 MeV
D) 32.4 MeV
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Answer:
Correct Answer: A
Solution:
- $ _1H^{2}+ _1H^{1}\to 2H _c^{4} $ initiate
$ \Rightarrow $ 1.1 × 4 = 4.4 final
$ \Rightarrow $ 4 × 7 = 28 release
$ \Rightarrow $ 28 - 4.4 = 23.6
