A beryllium-8 nucleus at rest undergoes double alpha decay as follows: $\begin{array} { l } 8 \\4\end{array}$ Be ? $\begin{array} { l } 4 \\2\end{array}$ He + $\begin{array} { l } 4 \\2\end{array}$ He The known atomic masses are: $\begin{array} { l } 4 \\2\end{array}$ He: 4.002603 u $\begin{array} { l } 8 \\4\end{array}$ Be: 8.005305 u
What is the kinetic energy of each departing alpha particle? (1 u = 931.5 MeV/c²)

Question

A beryllium-8 nucleus at rest undergoes double alpha decay as follows:

\begin{array} { l } 8 \\4\end{array}

Be ?

\begin{array} { l } 4 \\2\end{array}

He +

\begin{array} { l } 4 \\2\end{array}

He The known atomic masses are:

\begin{array} { l } 4 \\2\end{array}

He: 4.002603 u

\begin{array} { l } 8 \\4\end{array}

Be: 8.005305 u
What is the kinetic energy of each departing alpha particle? (1 u = 931.5 MeV/c²)

Quizplus · Accepted Answer

The correct answer is A.The total kinetic energy released in the decay is:$$Q = (8.005305 u - 2 	imes 4.002603 u) 	imes 931.5 MeV/c^2 = 92 keV$$Since the two alpha particles share the energy equally, each alpha particle has a kinetic energy of 46 keV.

A Beryllium-8 Nucleus at Rest Undergoes Double Alpha Decay as Follows