Question

Consider the nuclear reaction $A+B\to C$, and nuclei $A$ and $B$ are moving with kinetic energies of 5 $MeV$ and 3 $MeV$ respectively to form a nucleus $C$, which is in its excited state with excitation energy 10 $MeV$. If the $K.E$ of $C$ just after formation is $xMeV$, then the value of $x$ is
​​​​​​​[Answer upto two decimal points]
[Take mass of nuclei $A$, $B$ and $C$ are 25 $Da$, 10 $Da$ and 34.995 $Da$ respectively and $1Da=\frac{930MeV}{c^2}$]

Answer 1

Let $E$ be the excitation energy of nucleus $C$.
On conserving total energy,
$\Rightarrow T.E_A+T.E_B=T.E_C$
i.e., $\Rightarrow m_A{c}^2+K.E_A+m_B{c}^2+K.E_B=m_c{c}^2+K.E_C+E\Rightarrow (m_A+m_B-m_C)c^2+(8-10)MeV=K.E_C\Rightarrow K.E_C=(25+10-34.995)\frac{930}{c^2}.c^{2}MeV-2MeV=(4.65-2)MeV=2.65MeV$