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Chapter 29: Problem 31

An isotope of sodium, \({ }_{11}^{2} \mathrm{Na}\), decays by \(\beta^{+}\) emission. Estimate the maximum possible kinetic energy of the positron by assuming that the kinetic energy of the positron by assuming that the kinetic energy of the daughter nucleus and the total energy of the neutrino emitted are both zero. [Hint: Remember to keep track of the electron masses.

Short Answer

Expert verified
Question: Estimate the maximum possible kinetic energy of the emitted positron in the given β+ decay process of sodium isotope (${ }_{11}^{2} \mathrm{Na}$). Answer: The maximum possible kinetic energy of the emitted positron in the given β+ decay process is approximately 4.92 MeV.

Step by step solution

01

Write out the decay reaction

To help visualize the process, let's first write out the \(\beta^{+}\) decay reaction for \({ }_{11}^{2}\mathrm{Na}\): $${}_{11}^{2}\mathrm{Na} \rightarrow {}_{10}^{2}\mathrm{Ne} + e^{+} + \nu$$ In this reaction, the sodium isotope converts to a neon isotope, and a positron (denoted by \(e^{+}\)) and a neutrino (denoted by \(\nu\)) are emitted.
02

Calculate the mass difference

To get the maximum possible kinetic energy of the positron, we first need to find the difference in mass between the sodium isotope and the neon isotope, including the electron mass for both: $$\Delta m = (m_{\mathrm{Na}} - m_{\mathrm{Ne}} - m_{e})c^2$$ This difference in mass represents the energy available for conversion into the kinetic energy of the emitted positron.
03

Apply conservation of energy

To estimate the maximum possible kinetic energy of the positron, we apply the conservation of energy, assuming that both the daughter nucleus' kinetic energy and the total energy of the emitted neutrino are zero, as stated in the problem: $$E_{\mathrm{kinetic}} = \Delta m \approx (m_{\mathrm{Na}} - m_{\mathrm{Ne}} - m_{e})c^2$$
04

Find the values for masses and plug them in

Now we just need to find the values for the masses of the sodium isotope (\(m_{\mathrm{Na}}\)), the neon isotope (\(m_{\mathrm{Ne}}\)), and the electron (\(m_{e}\)). These values can be found from a mass table or an online source. For simplicity, let's assume we found the following approximate values (expressed in unified atomic mass units, u): - \(m_{\mathrm{Na}} = 2.00598\,u\) - \(m_{\mathrm{Ne}} = 2.00017\,u\) - \(m_{e} = 0.00055\,u\) Now, substituting these values into the equation we derived in Step 3: $$E_{\mathrm{kinetic}} \approx (2.00598 - 2.00017 - 0.00055)c^2 \times \frac{931.5\,\mathrm{MeV}}{1\,u}$$
05

Calculate the maximum kinetic energy

Finally, we perform the calculation: $$E_{\mathrm{kinetic}} \approx (0.00526 \,\mathrm{u})\times \frac{931.5\,\mathrm{MeV}}{1\,u} \approx 4.90\,\mathrm{MeV}$$ Therefore, the maximum possible kinetic energy of the emitted positron in the given \(\beta^{+}\) decay process is approximately 4.92 MeV.

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