Question

In Column I some of the nuclear reactions are given. Match this with the energy involved in these reactions in Column II :

Answer 1

Energy released = $△m{c}^2$

$A$ ${}_1^{2}H{+}_1^{2}H{\to }_1^{3}H{+}_1^{1}H+E_1△m=2m{(}_1^{2}H)-m{(}_1^{3}H)-m{(}_1^{1}H)=2\times 2.014-3.016-1.008=0.004amu{E}_1=△m{c}^2=0.004\times 931=3.7MeV\approx 4MeV$

So,$\left(A\right)\to \left(3\right)$

$B$ ${}_1^{3}H{+}_1^{2}H{\to }_2^{4}He{+}_0^{1}H+E_2△m=m{(}_1^{3}H)+m{(}_1^{2}H)-m{(}_2^{4}He)-m{(}_0^{1}H)=3.016+2.014-4.002-1.008=0.02amu{E}_2=△m{c}^2=0.02\times 931=18.62MeV$

So,$\left(B\right)\to \left(2\right)$

$C$ ${}_1^{2}H{+}_1^{2}H{\to }_2^{3}He{+}_0^{1}H+E_3△m=2m{(}_1^{2}H)-m{(}_2^{3}He)-m{(}_0^{1}H)=2\times 2.014-3.016-1.008=0.004amu{E}_3=0.004\times 931=3.3MeV$

So,$\left(C\right)\to \left(1\right)$

$D$ ${}_2^{3}H{+}_1^{2}H{\to }_2^{4}He{+}_1^{1}H+E_4△m=m{(}_2^{3}H)+m{(}_1^{2}H)-m{(}_2^{4}He)-m{(}_1^{1}H)=3.016+2.014-4.002-1.008=0.02amu{E}_4=△m{c}^2=0.02\times 931=18.62MeV$

So,$\left(D\right)\to \left(2\right)$