Question

The binding energies per nucleon for deuterium and helium are $1.1MeV$ and $7.0MeV$ respectively. The energy in joules that will be liberated when ${10}^6$ deuterons take part in the reaction

• A. $18.88\times {10}^{-3}J$
• B. $18.88\times {10}^{-5}J$
• C. $18.88\times {10}^{-7}J$
• D. $18.88\times {10}^{-10}J$

Answer 1

The correct option is C $18.88\times {10}^{-7}J$

Avg $BE{)}_D=1.1MeV$ Avg$BE{)}_{He}=7MeV$
$2_1^{2}H{\to }_2^{4}He$

$Q=BE{)}_{He}-2BE{)}_D$
$=7\times 4-2\times 1.1\times 2$
$=28-4.4$
$=23.6MeV$

For $2$ deuterons $23.6MeV$ is liberated.
For ${10}^6$, Energy liberated $=\frac{23.6}{2}\times {10}^6$
$=11.8\times {10}^6$ $MeV$
$1MeV=1.6\times {10}^{-13}J$
Energy liberated $=11.8\times {10}^6\times 1.6\times {10}^{-13}J$
$=18.88\times {10}^{-7}J$