NEET Solved Paper 2014 Question 42
Question: 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+ _2^{4}He+Q $ , the value of energy Q released is [AIPMT 2014]
Options:
A) 19.6 MeV
B) -2.4 MeV
C) 8.4 MeV
D) 17.3 MeV
Show Answer
Answer:
Correct Answer: D
Solution:
The binding energy for $ _1{H^{1}} $ is around zero and also not given in the question so we can ignore it
$ Q=24(4\times 7.06)-7\times (5.60) $ $ =(8\times 7.06)-(7\times 5.60) $ $ =(56.48-39.2)MeV $ $ =17.28MeV\simeq 17\text{.3MeV} $
BETA
