Question

Consider a fusion reaction ${}^4\text{He}{+}^4\text{He}{\to }^8\text{Be}.$ The $Q-$ value of the reaction is $-(90+n)\text{k eV}.$ The value of $n$ is (integer only).
Take $1\text{amu}=\frac{930}{c^2}\text{M eV}{\text{Atomic mass of}}^8\text{Be}=8.0053\text{u};{}^4\text{He}=4.0026\text{u}$

Answer 1

The given reaction can be written as,

${}^4\text{He}{+}^4\text{He}{\to }^8\text{Be}+Q$

$\Rightarrow Q=\Delta m{c}^2$

$\Rightarrow Q=\left[m{(}^4\text{He}{+}^4\text{He}\right)-m{(}^8\text{Be}\left)\right]\times c^2$

$=[2\times 4.0026-8.0053]\times c^2$

$=[-0.0001\text{amu}]\times c^2$

$=-0.093\text{M eV}\text{(or)}$

$=-93\text{k eV}=-(90+3)\text{k eV}$

$\therefore n=3$

Note : (Since $Q$ is negative, the fusion is not energetically favourable).