Question

The disintegration energy $Q$ for the change ${}_Z^A\text{X}\stackrel{-\alpha }{\to }{}_{Z-2}^{A-4}\text{D}+Q$ is given by $Q=\left[M{(}_Z^A\text{X}\right)-M{(}_{Z-2}^{A-4}\text{D})-M{(}_2^4\text{He})]C^2$, where $M$ represents the mass of respective nuclei reported in paranthesis and $C$ is speed of light. For $\alpha$-decay :

• A. $Q=+ve$
• B. $Q=-ve$
• C. $Q=0$
• D. not related with $Q$

Answer 1

The correct option is A $Q=+ve$

As in this reaction, mass decay occurs and this mass converts into energy according to Einstein as $E=\Delta m{C}^2=+ve$.