Radium-226 decays into radon-222 plus an alpha particle. The relevant masses are $\begin{array} { l } 4 \ 2 \end{array}$ He: 4.002603 u $\begin{array} { c } 222 \ 86 \end{array}$ Rn: 222.017570 u $\begin{array} { c } 226 \ 88 \end{array}$ Ra: 226.025402 u How much energy is released in this process? (1 u = 931.5 MeV/c²)

Question

Quizplus · Accepted Answer

The mass defect is calculated as: $ \Delta m = m_{\text{Ra}} - (m_{\text{Rn}} + m_{\text{He}}) = 226.025402 \, \text{u} - (222.017570 \, \text{u} + 4.002603 \, \text{u}) = 0.005229 \, \text{u} $. The energy released is $ \Delta E = \Delta m \times 931.5 \, \text{MeV/u} = 0.005229 \times 931.5 = 4.87 \, \text{MeV} $.

Radium-226 Decays into Radon-222 Plus an Alpha Particle 42\begin{array} { l } 4 \\2\end{array}42​

Radium-226 Decays into Radon-222 Plus an Alpha Particle $\begin{array} { l } 4 \\2\end{array}$