Question

The above is a plot of binding energy per nucleon $E_b$, against the nuclear mass $M;A,B,C,D,E,F$ correspond to different nuclei. Consider four reactions :
(i) $A+B\to C+ϵ$
(ii) $C\to A+B+ϵ$
(iii) $D+E\to F+ϵ$
(iv) $F\to D+E+ϵ$
where $ϵ$ is the energy released ? In which reaction is $ϵ$ positive ?

• A. (i) and (iv)
• B. (i) and (iii)
• C. (ii) and (iv)
• D. (ii) and (iii).

Answer 1

The correct option is A (i) and (iv)

The reactions for which product is/are more stable than reactant, energy & will be positive. (Energy will be released)

As B.E/nuclear decides stability

$\therefore$ Stability$:C>B$

$C>A$

$D>F$

and $E>F$

$\left(i\right)A+B⟶C+\epsilon$

$\left(iv\right)F⟶D+E+\epsilon$

are the reactions for which $\epsilon$ will be positive