Question

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.

Short answer

#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.

Step-by-step solution

Calculate the energy difference between states

Use the energy formula \(E_n = \frac{n^2h^2}{8mL^2}\) to find the energy of the ground state (n=1) and the first excited state (n=2): Ground state: \(E_1 = \frac{(1)^2 (6.626 \times 10^{-34} J \cdot s)^2}{8 (1.673 \times 10^{-27} kg) (10^{-14} m)^2}\) First excited state: \(E_2 = \frac{(2)^2 (6.626 \times 10^{-34} J \cdot s)^2}{8 (1.673 \times 10^{-27} kg) (10^{-14} m)^2}\) Calculate the energy difference: \(\Delta E = E_2 - E_1\)

Find the wavelength and frequency of the emitted photon

Use the energy difference to find the wavelength and frequency of the emitted photon. We know that the energy is conserved, so: \(\Delta E = h \nu \rightarrow \nu = \frac{\Delta E}{h}\) And we know that \(c = \nu \cdot \lambda\), so: \(\lambda = \frac{c}{\nu}\) Here c is the speed of light (3 * 10^8 m/s).

Find the spectrum range

Based on the wavelength we found, we can determine the spectrum range of the emitted photon. We can compare the wavelength with the well-known spectrum ranges (gamma, X-ray, ultraviolet, visible, infrared, microwave, radio waves).

Sketch the wave function for ground and excited states

Use the formula \(\psi_n(x) = \sqrt{\frac{2}{L}}\sin(\frac{n\pi x}{L})\) to sketch the wave function for the ground state and each of the first three excited states: 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): \(\psi_4(x) = \sqrt{\frac{2}{10^{-14} m}}\sin(\frac{4\pi x}{10^{-14} m})\) Sketch the wave functions on the same graph for a clear representation.