Question

Calculate the maximum kinetic energy of the \(\beta\) particle when $_{19}^{40} \mathrm{K}\( decays via \)\beta^{-}$ decay.

Short answer

Answer: The maximum kinetic energy of the beta particle during the radioactive decay of potassium-40 via beta decay is approximately 1.31 MeV.

Step-by-step solution

Write down the decay equation for the given process

The given radioactive decay process is the \(\beta^{-}\) decay of potassium-40 (\(_{19}^{40}\mathrm{K}\)). During \(\beta^{-}\) decay, a neutron in the nucleus is converted to a proton, producing an electron and an electron antineutrino. Mathematically, the decay equation for the \(\beta^{-}\) decay can be written as: n \(\rightarrow\) p + e\(^-\) + \(\bar{\nu}_e\) Applying this to \(_{19}^{40}\mathrm{K}\), we get: \(_{19}^{40}\mathrm{K} \rightarrow {_{20}^{40}\mathrm{Ca}} + e^- + \bar{\nu}_e\)

Find the Q-value for the decay process

The Q-value represents the energy difference between the initial and final particles involved in the decay. It can be calculated using the following formula: Q = (Mass of parent nucleus - Mass of daughter nucleus) * c\(^2\) In this problem, the parent nucleus is \(_{19}^{40}\mathrm{K}\) and the daughter nucleus is \(_{20}^{40}\mathrm{Ca}\). The masses of these nuclei are: - \(_{19}^{40}\mathrm{K}\) = 39.963998166(28) u - \(_{20}^{40}\mathrm{Ca}\) = 39.962590863(50) u Where u is the atomic mass unit, 1 u = 931.494 MeV/c\(^2\). Now we can calculate the Q-value: Q = [(39.963998166 - 39.962590863) * 931.494] MeV Q ≈ 1.31 MeV

Calculate the maximum kinetic energy of the \(\beta\) particle

The maximum kinetic energy of the \(\beta\) particle occurs when the electron antineutrino carries away the minimum possible energy. Ideally, when the neutrino carries no energy, all of the available energy (Q) will be transferred to the kinetic energy of the \(\beta\) particle. Therefore, the maximum kinetic energy of the \(\beta\) particle will be equal to the Q-value: Kinetic energy of \(\beta\) particle = Q Kinetic energy of \(\beta\) particle = 1.31 MeV Thus, the maximum kinetic energy of the \(\beta\) particle during the \(_{19}^{40}\mathrm{K}\) decay via \(\beta^{-}\) decay is approximately 1.31 MeV.