Question

A nucleus with mass number $184$ initially at rest emits an $\alpha -\text{particle}$. If the $Q$ value of the reaction is $5.5\text{MeV},$ calculate the kinetic energy of the $\alpha -\text{particle}$.

• A. $5.0\text{MeV}$
• B. $5.5\text{MeV}$
• C. $0.12\text{MeV}$
• D. $5.38\text{MeV}$

Answer 1

The correct option is D $5.38\text{MeV}$


From the conservation of energy,

$\frac{1}{2}\left(4m\right)v^2+\frac{1}{2}\left(180m\right){\left(\frac{4v}{180}\right)}^2=5.5\text{MeV}$

$\Rightarrow \frac{1}{2}\left(4m\right)v^2[1+45{\left(\frac{4}{180}\right)}^2]=5.5\text{MeV}$

$\Rightarrow K.E_{\alpha }=\frac{5.5}{1+45{\left(\frac{4}{180}\right)}^2}\text{MeV}$

$\therefore K.E_{\alpha }=5.38\text{MeV}$

Hence, $\left(D\right)$ is the correct answer.