Meteorites created in the early solar system contaned aluminum-26, which is a radioactive isotope of aluminum with a half-life of \(7.2 \times 10^5 \mathrm{yrs}\). Aluminum-26 decays first into an excited state of magnesium-26 via the reaction \({ }_{13}^{26} \mathrm{Al} \rightarrow{ }_{12}^{26} \mathrm{Mg}^*+\mathrm{e}^{+}\), where the \(e^{+}\)has energy \(2.99 \mathrm{MeV}\). (The \(e^{+}\)is a positron; see previous problem. The asterisk (*) indicates "excited.") The \({ }_{12}^{26} \mathrm{Mg}^*\) then decays into the stable isotope magnesium- 26 via the reaction \({ }_{12}^{26} \mathrm{Mg}^* \rightarrow{ }_{12}^{26} \mathrm{Mg}+\gamma\). The \(\gamma\) has energy \(1.8 \mathrm{MeV}\). a) If you were asked to calculate the de Broglie wavelength of the positron, would it be permissible to use Newtonian physics? Justify your answer. b) What is the wavelength of the photon emitted when the excited magnesium-26 decays into its ground state? What is its momentum? c) \({ }_{12}^{26} \mathrm{Mg}\) has an atomic mass of \(25.9826 \mathrm{u}\). What is the speed of the recoiling nucleus when the photon is emitted? d) What is the nucleus' kinetic energy in electron volts? e) Precise measurements indicate that for a certain meteorite \(A\) the present ratio \({ }^{26} \mathrm{Mg} /{ }^{27} \mathrm{Al}=5 \times 10^{-5}\), where \({ }^{27} \mathrm{Al}\) is the common, stable isotope of aluminum. For a meteorite \(B\) the ratio is \({ }^{26} \mathrm{Mg} /{ }^{27} \mathrm{Al}=1.55 \times 10^{-7}\). Assuming that the different ratios are due to the difference in the meteorites' times of creation, how much older is meteorite \(B\) than \(A\) ?

First, let's determine whether we can use Newtonian physics to calculate the de Broglie wavelength of the positron. To answer this question, we need to compare the Classical (Newtonian) limit to the relativistic (quantum) limit. The emitted positron has energy \(E = 2.99 \; \mathrm{MeV}\). We will first convert this energy into Joules: \(E = 2.99 \times 10^6 \; \mathrm{eV} \times 1.6 \times 10^{-19} \; \mathrm{J/\textit{eV}} = 4.784 \times 10^{-13} \; \mathrm{J}\). Now, we need to find the momentum \(p\) of the positron using the energy-momentum formula: \(E^2 = (mc^2)^2 + (pc)^2\), where \(m\) is the mass of the positron and \(c\) is the speed of light. By reorganizing the formula, we get: \(p = \sqrt{(E/c)^2 - (mc)^2} = \sqrt{(4.784 \times 10^{-13} \mathrm{J} / 3 \times 10^8 \mathrm{m/s})^2 - (9.109 \times 10^{-31} \mathrm{kg} \times 3 \times 10^8 \mathrm{m/s})^2} = 2.543 \times 10^{-22} \mathrm{kg~m/s}\). The de Broglie wavelength of the positron is given by \(\lambda = h/p\), where \(h\) is the Planck's constant. However, before calculating the wavelength, we need to check if the positron's speed is non-relativistic, meaning \(v \ll c\). In this case, \(E \approx pc\), and we can assume \(v = E/p\). \(v = E/p = (4.784 \times 10^{-13} \mathrm{J})/(2.543 \times 10^{-22} \mathrm{kg~m/s}) = 1.882 \times 10^9 \mathrm{m/s}\). As \(v \approx 6.3c\), the positron's speed is relativistic, and we cannot use Newtonian physics to calculate its de Broglie wavelength.

The emitted photon has energy of \(1.8 \; \mathrm{MeV}\). To find the wavelength of the photon, we use the following formula: \(E = h \nu\), where \(E\) is the energy, \(h\) is the Planck's constant and \(\nu\) is the frequency of the light. We have to convert the energy into Joules: \(E = 1.8 \times 10^6 \; \mathrm{eV} \times 1.6 \times 10^{-19} \; \mathrm{J/\textit{eV}} = 2.88 \times 10^{-13} \; \mathrm{J}\). Now, we will find the frequency of the light: \(\nu = E/h = (2.88 \times 10^{-13} \mathrm{J})/(6.626 \times 10^{-34} \mathrm{J~s}) = 4.345 \times 10^{20} \; \mathrm{Hz}\). The wavelength of the photon is given by \(\lambda = c / \nu\): \(\lambda = (3 \times 10^8 \;\mathrm{m/s})/(4.345 \times 10^{20} \; \mathrm{Hz}) = 6.906 \times 10^{-13} \; \mathrm{m}\). Now, we will find the momentum of the photon \(p\). Since \(E = pc\), we have: \(p = E/c = (2.88 \times 10^{-13} \;\mathrm{J})/(3 \times 10^8 \; \mathrm{m/s}) = 9.6 \times 10^{-22} \; \mathrm{kg~m/s}\).

We are supposed to find the speed of the recoiling magnesium-26 nucleus after the decay. To do this, we will apply conservation of momentum. The initial momentum of the system, including the excited magnesium-26 nucleus and the emitted photon, is zero. Therefore, the final momentum of the recoiling nucleus must equal the momentum of the photon, which is \(9.6 \times 10^{-22} \; \mathrm{kg~m/s}\). First, we need to calculate the mass of the magnesium-26 nucleus. The mass of \({}_{12}^{26} \mathrm{Mg}\) is given as \(25.9826 \; \mathrm{u}\). We convert this mass to kilograms: \(m = 25.9826 \;\mathrm{u} \times 1.66 \times 10^{-27} \;\mathrm{kg/\textit{u}} = 4.313 \times 10^{-26} \;\mathrm{kg}\). Finally, we can find the speed of the recoiling nucleus by dividing its momentum by its mass: \(v = p/m = (9.6 \times 10^{-22} \;\mathrm{kg~m/s})/(4.313 \times 10^{-26} \; \mathrm{kg}) = 2.225 \times 10^4 \; \mathrm{m/s}\).

We know the mass and speed of the recoiling nucleus. Hence, we can find its kinetic energy using the following formula: \(KE = \frac{1}{2}mv^2\). Don't forget, the result should be expressed in electron volts. \(KE = \frac{1}{2} \times (4.313 \times 10^{-26} \;\mathrm{kg}) \times (2.225 \times 10^4 \; \mathrm{m/s})^2 = 1.067 \times 10^{-21} \mathrm{J}\). Now we convert the energy to electron volts: \(KE = 1.067 \times 10^{-21} \;\mathrm{J} \times (6.242 \times 10^{18} \;\mathrm{eV/J}) = 6.66 \times 10^3 \; \mathrm{eV}\).

In this part of the exercise, we are given the present \({}^{26}\mathrm{Mg}/{ }^{27}\mathrm{Al}\) ratios for meteorites \(A\) and \(B\), which are \(5 \times 10^{-5}\) and \(1.55 \times 10^{-7}\), respectively. We assume that these different ratios are due to the different ages of the meteorites. We will use the decay formula \(N(t) = N_0 \exp{(-\lambda t)}\), where \(N(t)\) is the number of radioactive atoms remaining at time \(t\), \(N_0\) is the initial number of radioactive atoms, \(\lambda\) is the decay constant, and \(t\) is time. Since both meteorites started with the same \({}^{26}\mathrm{Mg}/{ }^{27}\mathrm{Al}\) ratio, we can denote the initial ratio as \(R_0\), and write the decay equations for meteorite \(A\) and \(B\): $R_A = R_0 \exp{(-\lambda t_A)} \\ R_B = R_0 \exp{(-\lambda t_B)}\(. We are supposed to find the time difference, meaning \)t_B - t_A$. From these equations, we can find the ratio between the ratios, this gives: \(\frac{R_A}{R_B} = \frac{\exp{(-\lambda t_A)}}{\exp{(-\lambda t_B)}} = \exp{(\lambda (t_B - t_A))}\). Taking the natural logarithm of both sides, we get: \(\lambda (t_B - t_A) = \ln{\left(\frac{R_A}{R_B}\right)}\). Now we find the half-life of aluminum-26: \(T_{1/2} = 7.2 \times 10^5 \;\mathrm{yrs}\). Using the relationship between the decay constant and half-life, we find \(\lambda = \frac{\ln{2}}{T_{1/2}} = 9.63 \times 10^{-7} \; \mathrm{yr}^{-1}\). Finally, we can find the age difference between the meteorites using the ratio of ratios: \(t_B - t_A = \frac{\ln{\left(\frac{R_A}{R_B}\right)}}{\lambda} = \frac{\ln{\left(\frac{5 \times 10^{-5}}{1.55 \times 10^{-7}}\right)}}{9.63 \times 10^{-7} \;\mathrm{yr}^{-1}} = 1.63 \times 10^6 \;\mathrm{yrs}\). Hence, meteorite \(B\) is older than meteorite \(A\) by \(1.63 \times 10^6\) years.