Nuclei - Result Question 35
36. In the reaction, $ _1^{2} H+ _1^{3} H \longrightarrow _2^{4} He+ _0^{1} n$, if the binding energies of $ _1^{2} H, _1^{3} H$ and $ _2^{4} He$ are respectively, $a, b$ and $c$ (in $MeV$ ), then the energy (in $MeV$ ) released in this reaction is
(a) $a+b+c$
(b) $a+b-c$
(c) $c-a-b$
(d) $c+a-b$
[2005]
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Answer:
Correct Answer: 36. (c)
Solution:
- (c) $ _1^{2} H$ and $ _1^{3} H$ requires $a$ and $b$ amount of energies for their nucleons to be separated.
$ _2^{4} He$ releases $c$ amount of energy in its formation i.e., in assembling the nucleons as nucleus.
Hence, Energy released $=c-(a+b)=c-a-b$
