Question

The Binding energy per nucleon of ${}_3^{7}Li$ and ${}_2^{4}He$ nucleon are $5.60MeV$ and $7.06MeV$, 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. $19.6MeV$
• B. $-2.4MeV$
• C. $8.4MeV$
• D. $17.3MeV$

Answer 1

Answer: (D)

$17.3MeV$

${}_{4}L{i}^7{+}_1{H}^1{\to }_2{(}_{2}H{e}^4)$

BE of products $=\left(5.6MeV\right)\times 7+0$$=39.2MeV$
$E=-39.2MeV$

BE of reactant $=\left(7.06\right)\times 4\times 2$$=56.48MeV$
$E_f=-56.48MeV$

As nuclear energy decreases, so some energy will be released
$Q_{release}=E_i-E_f=(-39.2)-(-56.48)=17.28MeV$