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 :
$$\left( {\text{i}} \right)\,A + B \to C + \varepsilon $$
$$\left( {{\text{ii}}} \right)\,C \to A + B + \varepsilon $$
$$\left( {{\text{iii}}} \right)\,D + E \to F + \varepsilon \,{\text{and}}$$
$$\left( {{\text{iv}}} \right)\,F \to D + E + \varepsilon ,$$
where $$\varepsilon $$ is the energy released? In which reactions is $$\varepsilon $$ positive?
A.
(i) and (iii)
B.
(ii) and (iv)
C.
(ii) and (iii)
D.
(i) and (iv)
Answer :
(i) and (iv)
Solution :
For $$A + B \to C + \varepsilon ,\varepsilon $$ is positive. This is because $${E_b}$$ for $$C$$ is greater than the $${E_b}$$ for $$A$$ and $$B.$$
Again for $$F \to D + E + \varepsilon ,\varepsilon $$ is positive. This is because $${E_b}$$ for $$D$$ and $$E$$ is greater than $${E_b}$$ for $$F.$$
For $$A + B \to C + \varepsilon ,\varepsilon $$ is positive. This is because $${E_b}$$ for $$C$$ is greater than the $${E_b}$$ for $$A$$ and $$B.$$
Again for $$F \to D + E + \varepsilon ,\varepsilon $$ is positive. This is because $${E_b}$$ for $$D$$ and $$E$$ is greater than $${E_b}$$ for $$F.$$