Nuclear Physics And Radioactivity Question 97
Question: The binding energy of deuteron $ \begin{matrix} 2 \ 1 \ \end{matrix}H $ is 1.112 Me V pen nucleon and an $ \alpha $ -particle $ \begin{matrix} 4 \ 2 \ \end{matrix}He $ has a binding energy of 7.047 Me V per nucleon. Then in the fusion reaction $ \begin{matrix} 2 \ 1 \ \end{matrix}He+\begin{matrix} 2 \ 1 \ \end{matrix}H\to \begin{matrix} 4 \ 2 \ \end{matrix}He+Q $ , the energy Q released is
Options:
A) $ 1,MeV $
B) $ 11.9,MeV $
C) $ 23.8,MeV $
D) $ 931,MeV $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Mass of $ _{1}H^{2} $ = 2.01478 a.m.u. Mass of $ _{2}H^{4} $ = 4.00388 a.m.u. Mass of two deuterium $ =2\times 2.01478=4.02956 $
Energy equivalent to $ 2_{1}H^{2} $
$ =4.02956\times 1.112 $ MeV $ =4.48 $ MeV Energy equivalent to $ _{2}H^{4} $
$ =7.00388\times 7.047 $ MeV $ =28.21MeV $ Energy released $ =28.21-4.48 $ $ =23.73 $ MeV $ =24 $ MeV
BETA
