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

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

Answer 1

The correct option is D 17.3 eV

Q=2(BE of He)-(BE of Li)
$=2\times (4\times 7.06)-(7\times 5.06)$
=56.48-39.2=17.3 MeV