\(^{249} \mathrm{Cf}\) undergoes alpha decay. (a) Write the reaction equation. (b) Find the energy released in the decay.

In alpha decay, the unstable heavy nucleus emits an alpha particle (consisting of 2 protons and 2 neutrons) and results in a new nucleus. The general reaction equation for alpha decay is: \(A_{initial}\mathrm{X} \rightarrow A_{final}\mathrm{Y} + ^4_2\mathrm{He}\), where \(A_{initial}\) and \(A_{final}\) are the mass numbers of the initial and final nuclei, and X and Y are the symbols of the initial and final elements, respectively. In our case, the Californium-249 (\(^{249}\mathrm{Cf}\)) undergoes alpha decay. The mass number of the final nucleus is 249 - 4 = 245. The proton number (atomic number) decreases by 2. Since Californium has an atomic number of 98, the final nucleus will have an atomic number of 96, which corresponds to the element Curium (Cm). Therefore, the reaction equation for the alpha decay of Californium-249 is: \(^{249}\mathrm{Cf} \rightarrow ^{245}\mathrm{Cm} + ^4_2\mathrm{He}\).

To find the energy released during the decay, we can use the binding energy per nucleon curve and the mass defect. The energy released can be expressed as follows: \(E = \Delta m \times c^2\), where \(E\) is the energy released, \(\Delta m\) is the mass defect, and \(c\) is the speed of light in a vacuum. The mass defect represents the difference in mass before and after an alpha decay process. We can calculate the mass defect as follows: \(\Delta m = m_{initial} - m_{final} - m_{alpha}\), where \(m_{initial}\), \(m_{final}\), and \(m_{alpha}\) are the masses of the initial nucleus, final nucleus, and alpha particle, respectively. The mass of an alpha particle is approximately 4 atomic mass units (amu). Using the atomic mass unit data, we can approximate the mass defect: \(\Delta m \approx (^{249}\mathrm{Cf}) - (^{245}\mathrm{Cm}) - 4 \,\mathrm{amu}\). Note that one atomic mass unit (amu) is approximately equal to \(1.6605 \times 10^{-27}\, kg\). Now we can calculate the mass defect in kilograms: \(\Delta m \approx (\,^{249}\mathrm{Cf} - \,^{245}\mathrm{Cm} - 4)\times 1.6605 \times 10^{-27}\, kg\). Finally, using the speed of light, c = \(2.998\times 10^8\, m/s\), we can calculate the energy released: \(E = \Delta m \times c^2\). This energy, expressed in joules, can be converted to more familiar units like mega electron volts (MeV) by dividing the result by \(1.602\times 10^{-13}\, J/MeV\). The final step involves looking up the atomic masses of the Californium-249 and Curium-245 isotopes (usually given in an atomic mass unit table), completing the mass defect and energy released calculations, and reporting the energy in MeV.