MP Board · Class 12 · Physics · NucleiCalculate the energy released in the alpha decay of ${}{88}^{226}\text{Ra}$ into ${}{86}^{222}\text{Rn}$, given the following atomic masses: Mass of ${}{88}^{226}\text{Ra} = 226.02540\text{ u}$ Mass of ${}{86}^{222}\text{Rn} = 222.01757\text{ u}$ Mass of ${}{2}^{4}\text{He} = 4.00260\text{ u}$\nUse the conversion factor $1\text{ u} = 931.5\text{ MeV}/c^2$. Also write the nuclear reaction equation for this decay.

Step 1: Write the Nuclear Decay Equation\nThe alpha decay of radium-226 into radon-222 is represented by the following equation:

$${}{88}^{226}\text{Ra} \rightarrow {}{86}^{222}\text{Rn} + {}_{2}^{4}\text{He}$$

Step 2: Calculate the Mass Defect ($\Delta m$)\nMass defect is the difference between the total mass of the reactants (parent nucleus) and the total mass of the products (daughter nucleus + alpha particle).

• Mass of parent nucleus (${}_{88}^{226}\text{Ra}$), $m_p = 226.02540\text{ u}$
• Mass of daughter nucleus (${}_{86}^{222}\text{Rn}$), $m_d = 222.01757\text{ u}$
• Mass of alpha particle (${}{2}^{4}\text{He}$), $m{\alpha} = 4.00260\text{ u}$ \nTotal mass of products = $m_d + m_{\alpha}$ $$\text{Total mass of products} = 222.01757\text{ u} + 4.00260\text{ u} = 222.02017\text{ u}$$ \nNow, calculate the mass defect $\Delta m$: $$\Delta m = m_p - (m_d + m_{\alpha})$$ $$\Delta m = 226.02540\text{ u} - 222.02017\text{ u}$$ $$\Delta m = 0.00523\text{ u}$$

Step 3: Calculate the Energy Released ($Q$)\nThe energy released in the reaction is given by:

$$Q = \Delta m \times 931.5\text{ MeV}$$ \nSubstitute the value of $\Delta m$: $$Q = 0.00523\text{ u} \times 931.5\text{ MeV}/\text{u}$$ $$Q = 4.8717\text{ MeV}$$

Conclusion\nThe energy released during the alpha decay of ${}_{88}^{226}\text{Ra}$ is approximately $4.87\text{ MeV}$.