The particle in a box model is often used to make rough estimates of energy level spacings. Suppose that you have a proton confined to a one-dimensional box of length equal to a nuclear diameter (about \(10^{-14} \mathrm{m}\) ). (a) What is the energy difference between the first excited state and the ground state of this proton in the box? (b) If this energy is emitted as a photon as the excited proton falls back to the ground state, what is the wavelength and frequency of the electromagnetic wave emitted? In what part of the spectrum does it lie? (c) Sketch the wave function \(\psi\) as a function of position for the proton in this box for the ground state and each of the first three excited states.

#Answer# (a) The energy difference between the first excited state and the ground state of the proton is $\Delta E ≈ 9.78 × 10^{-19} J$. (b) The wavelength of the electromagnetic wave emitted is $\lambda ≈ 2.01 × 10^{-14} m$, and its frequency is $\nu ≈ 1.50 × 10^{19} Hz$. This falls within the gamma-ray part of the spectrum. (c) The wave functions for the ground state and each of the first three excited states are: - Ground state (n=1): $\psi_1(x) = \sqrt{\frac{2}{10^{-14} m}}\sin(\frac{1\pi x}{10^{-14} m})$ - First excited state (n=2): $\psi_2(x) = \sqrt{\frac{2}{10^{-14} m}}\sin(\frac{2\pi x}{10^{-14} m})$ - Second excited state (n=3): $\psi_3(x) = \sqrt{\frac{2}{10^{-14} m}}\sin(\frac{3\pi x}{10^{-14} m})$ - Third excited state (n=4): $\psi_4(x) = \sqrt{\frac{2}{10^{-14} m}}\sin(\frac{4\pi x}{10^{-14} m})$ The sketch of the wave functions shows that as the quantum number n increases, the number of nodes in the wave function also increases, which indicates the increased energy of the excited states.