When a stationary plutonium-239, $\begin{array} { c } 239 \ 94 \end{array}$ Pu, decays into uranium-235 plus an alpha particle, the energy released in the process is 5.24 MeV. The following masses are known: $\begin{array} { l } 4 \ 2 \end{array}$ He: 4.002603 u $\begin{array} { c } 235 \ 92 \end{array}$ U: 235.043924 u What is the mass of the $\begin{array} { c } 239 \ 94 \end{array}$ Pu nucleus, in amu? (1 u = 931.494 MeV/c²)

Question

Quizplus · Accepted Answer

The mass of the $\begin{array} { c } 239 \94\end{array}$ Pu nucleus is equal to the sum of the masses of the $\begin{array} { c } 235 \92\end{array}$ U nucleus and the alpha particle, minus the energy released in the process. The energy released in the process is 5.24 MeV, which is equivalent to 5.24 MeV/c² = 5.61 x 10^-3 u. Therefore, the mass of the $\begin{array} { c } 239 \94\end{array}$ Pu nucleus is 235.043924 u + 4.002603 u - 5.61 x 10^-3 u = 239.05215 u.

When a Stationary Plutonium-239 23994\begin{array} { c } 239 \\94\end{array}23994​ Pu, Decays into Uranium-235 Plus an Alpha Particle, the Energy

When a Stationary Plutonium-239 $\begin{array} { c } 239 \\94\end{array}$ Pu, Decays into Uranium-235 Plus an Alpha Particle, the Energy