Nuclei Ques 19
19. The Binding energy per nucleon of $ _3^{7} Li$ and $ _2^{4} He$ nuclei are $5.60 MeV$ and $7.06 MeV$, respectively. In the nuclear reaction $ _3^{7} Li+ _1^{1} H \to _2^{4} He+Q$, the value of energy $Q$ released is :
[2014]
(a) $19.6 MeV$
(b) $-2.4 MeV$
(c) $8.4 MeV$
(d) $17.3 MeV$
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Answer:
Correct Answer: 19.(d)
Solution:
- (d) $BE$ of $ _2 He^{4}=4 \times 7.06=28.24 MeV$
$BE$ of $ _3^{7} Li=7 \times 5.60=39.20 MeV$
$ _3^{7} Li+ _1^{1} H \to _2 He^{4}+ _2 He^{4}+Q$
$39.20 \qquad \qquad 28.24\times 2(=56.48 MeV) $
Therefore, $Q = 56.48-39.20=17.28 MeV$
BETA
