Question

The binding energy per nucleon of $Li$ and $He$ nuclei are $5.60\, MeV$ and $7.06\, MeV$ respectively. In the nuclear reaction $^7_3Li+_1^1H \rightarrow _2^4He + _2^4He+Q$ the value of energy Q released is

• A. 19.6 MeV
• B. -2.4 MeV
• C. 8.4 MeV
• D. 17.3 MeV

Answer 1

Answer: (D)

17.3 MeV

Answer 2

The correct answer is D:17.3MeV
Binding energy of $^7_3Li$ nucleus
=7 $\times 5.60 MeV=39.2 MeV$
Binding energy of $_2^4 He$nucles
=4 $\times 7.06 MeV=28.24 MeV$
The reaction is
$^7_3Li + _1^1H \rightarrow \, 2(_2^4 He)+Q$
$\therefore Q=2(BE \space of\space _2^4He)-(BE \, of \space _3^7Li)$
=2 $\times$28.24 MeV-39.2 MeV
=56.48 MeV-39.2 MeV
=17.28 MeV