Question

In the reaction ${}_1^2\text{H}{+}_1^3\text{H}{\to }_2^4\text{He}{+}_0^1\text{n}$. If the binding energies of ${}_1^2\text{H}{,}_1^3\text{H}$ and ${}_2^4\text{He}$ are respectively $a,b$ and $c$ (in MeV), then the energy(in MeV) released in this reaction is:

• A. $a+b+c$
• B. $c+a-b$
• C. $c-a-b$
• D. $a+b-c$

Answer 1

The correct option is C $c-a-b$

${}_1^{2}H{+}_1^{3}H{⟶}_2^{4}He{+}_0^{1}n$

Energy released, $E=\left(△m\right)\times 931MeV$

(Note:-$1$ atomic mass unit$=931MeV$)

$△m=$mass of product$-$ mass of reactant

$△m=c-a-b$

$E=\left(△m\right)\times 931$

or $E=(c-a-b)$

(The binding energies of ${}_1^{2}H{,}_1^{3}H$ and ${}_2^{4}He$ are respectively a,b and c)