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Chapter 29: Problem 27

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

Short Answer

Expert verified
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

01

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: \(_{19}^{40}\mathrm{K} \rightarrow {_{20}^{40}\mathrm{Ca}} + e^- + \bar{\nu}_e\)
02

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
03

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.

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Most popular questions from this chapter

Approximately what is the total energy of the neutrino emitted when ${ }_{11}^{22} \mathrm{Na}$ decays by electron capture?
A certain radioactive nuclide has a half-life of \(200.0 \mathrm{s}\) A sample containing just this one radioactive nuclide has an initial activity of \(80,000.0 \mathrm{s}^{-1} .\) (a) What is the activity 600.0 s later? (b) How many nuclei were there initially? (c) What is the probability per second that any one of the nuclei decays?
The nucleus in a \({ }_{7}^{12} \mathrm{~N}\) atom captures one of the atom's electrons, changing the nucleus to \({ }_{6}^{12} \mathrm{C}\) and emitting a neutrino. What is the total energy of the emitted neutrino? [Hint:You can use the classical expression for the kinetic energy of the ${ }_{6}^{12} \mathrm{C}$ atom and the extremely relativistic expression for the kinetic energy of the neutrino.]
The power supply for a pacemaker is a small amount of radioactive $^{238} \mathrm{Pu} .\( This nuclide decays by \)\alpha$ decay with a half-life of 86 yr. The pacemaker is typically replaced every 10.0 yr. (a) By what percentage does the activity of the \(^{238}\) Pu source decrease in 10 yr? (b) The energy of the \(\alpha\) particles emitted is 5.6 MeV. Assume an efficiency of \(100 \%\) -all of the \(\alpha\) -particle energy is used to run the pacemaker. If the pacemaker starts with \(1.0 \mathrm{mg}\) of \(^{238} \mathrm{Pu}\) what is the power output initially and after 10.0 yr?
In this problem, you will verify the statement (in Section 29.4) that the \(^{14} \mathrm{C}\) activity in a living sample is \(0.25 \mathrm{Bq}\) per gram of carbon. (a) What is the decay constant \(\lambda\) for \(^{14} \mathrm{C} ?\) (b) How many \(^{14} \mathrm{C}\) atoms are in \(1.00 \mathrm{g}\) of carbon? One mole of carbon atoms has a mass of \(12.011 \mathrm{g},\) and the relative abundance of \(^{14} \mathrm{C}\) is \(1.3 \times 10^{-12} .\) (c) Using your results from parts (a) and (b), calculate the \(^{14} \mathrm{C}\) activity per gram of carbon in a living sample.
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