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Before the discovery of the neutron, one theory of the nucleus proposed that the nucleus contains protons and electrons. For example, the helium-4 nucleus would contain 4 protons and 2 electrons instead of - as we now know to be true- 2 protons and 2 neutrons. (a) Assuming that the electron moves at nonrelativistic speeds, find the ground-state energy in mega-electron- volts of an electron confined to a one-dimensional box of length $5.0 \mathrm{fm}\( (the approximate diameter of the \)^{4} \mathrm{He}$ nucleus). (The electron actually does move at relativistic speeds. See Problem \(80 .)\) (b) What can you conclude about the electron-proton model of the nucleus? The binding energy of the \(^{4} \mathrm{He}\) nucleus - the energy that would have to be supplied to break the nucleus into its constituent particles-is about \(28 \mathrm{MeV} .\) (c) Repeat (a) for a neutron confined to the nucleus (instead of an electron). Compare your result with (a) and comment on the viability of the proton-neutron theory relative to the electron-proton theory.

Step by step solution

01

Ground-state energy of a particle in a one-dimensional box

First, we need to find a general equation for the ground-state energy of a particle confined in a one-dimensional box of length L. Using the Schrödinger equation and assuming that the potential barrier within the box is infinitely high, we can derive the following expression for the energy of the particle in the ground-state: $$E_n = \frac{n^2 \hbar^2 \pi^2}{2mL^2}$$ where \(E_n\) is the energy of the particle in the \(n\)-th energy level (n=1 for the ground-state), \(n\) is the quantum number, \(\hbar\) is the reduced Planck constant, \(m\) is the mass of the particle, and \(L\) is the length of the box. Since we are only interested in the ground-state energy, we can set n=1 and obtain the following formula: $$E_1 = \frac{\hbar^2 \pi^2}{2mL^2}$$

02

Ground-state energy of an electron confined to the nucleus

Now, we can use the formula derived in the previous step to find the ground-state energy of an electron confined to the nucleus. Knowing the values for \(\hbar = 1.054 \times 10^{-34} \mathrm{J\: s}\), the mass of an electron \(m_e = 9.109 \times 10^{-31} \mathrm{kg}\), and the box length \(L = 5.0 \times 10^{-15} \mathrm{m}\), we can calculate the ground-state energy of an electron: $$E_{1e} = \frac{(1.054 \times 10^{-34})^2 \pi^2}{2(9.109 \times 10^{-31})(5.0 \times 10^{-15})^2}$$ $$E_{1e} = 3.658 \times 10^{-14} \mathrm{J}$$ Converting this energy to mega-electron volts, we find the ground-state energy for the electron: $$E_{1e} = \frac{3.658 \times 10^{-14}}{1.602 \times 10^{-13}} \mathrm{MeV}$$ $$E_{1e} \approx 0.228 \mathrm{MeV}$$

03

Comparison with the helium-4 nucleus binding energy

The binding energy of the helium-4 nucleus is given as 28 MeV. Comparing this value with the ground-state energy of the electron confined to the nucleus (0.228 MeV), we can conclude that the electron-proton model is not a viable explanation for the nucleus. The binding energy supplied by an electron is vastly lower compared to the experimental value, which implies that there must be other particles in the nucleus that contribute to the binding energy.

04

Ground-state energy of a neutron confined to the nucleus

Next, let's calculate the ground-state energy of a neutron confined in the nucleus. For this, we need the mass of a neutron, \(m_n = 1.675 \times 10^{-27} \mathrm{kg}\). Using the expression obtained in step 1, we can find the ground-state energy for the neutron: $$E_{1n} = \frac{(1.054 \times 10^{-34})^2 \pi^2}{2(1.675 \times 10^{-27})(5.0 \times 10^{-15})^2}$$ $$E_{1n} = 1.469 \times 10^{-11} \mathrm{J}$$ Converting this energy to mega electron volts, we find the ground-state energy for the neutron: $$E_{1n} = \frac{1.469 \times 10^{-11}}{1.602 \times 10^{-13}} \mathrm{MeV}$$ $$E_{1n} \approx 9.168 \mathrm{MeV}$$

05

Comparison of the electron and neutron ground-state energies, and final conclusion

Comparing the ground-state energies for the electron (\(E_{1e} \approx 0.228 \mathrm{MeV}\)) and neutron (\(E_{1n} \approx 9.168 \mathrm{MeV}\)), we find that the contribution from a neutron is much higher, although still lower than experimental binding energies in the \(^{4}\mathrm{He}\) nucleus. However, the energy provided by two neutrons and two protons together would actually be reasonable to assume an alternative explanation to the electron-proton model. Therefore, when considering the viability of the proton-neutron model relative to the electron-proton model, the proton-neutron model is more consistent with experimental evidence, ultimately leading to the discovery of the neutron and the current understanding of atomic nuclei.

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