Question

Determine the three lowest energies of the wave function of a proton in a box of width \(1.0 \cdot 10^{-10} \mathrm{~m}\)

Short answer

Answer: The three lowest energies are \(E_1 \approx 1.53 \cdot 10^{-25} \mathrm{~J}\), \(E_2 \approx 6.12 \cdot 10^{-25} \mathrm{~J}\), and \(E_3 \approx 1.38 \cdot 10^{-24} \mathrm{~J}\).

Step-by-step solution

Identify the one-dimensional particle in a box energy equation

For a one-dimensional particle in a box scenario, the energy of the particle is given by the following equation: \(E_n = \dfrac{n^2 h^2}{8 m L^2}\) where \(E_n\) represents the energy of the particle at level n, n is the energy level (n=1, 2, 3, ...), \(h\) is the Planck's constant \((6.626 \cdot 10^{-34} \mathrm{~J\cdot s})\), m is the mass of the particle, and L is the width of the box.

Analyze and define the given scenario

The exercise specifies that the particle in the box is a proton, which has a mass of approximately \(1.67 \cdot 10^{-27} \mathrm{~kg}\). The width of the box is given as \(1.0 \cdot 10^{-10} \mathrm{~m}\). We need to calculate the lowest three energies for this scenario, which means we need to find \(E_1\), \(E_2\), and \(E_3\) using the energy equation provided in step 1.

Calculate the energy at level n=1

Plug in the values for n=1, m, and L into the energy equation and solve for \(E_1\): \(E_1 = \dfrac{1^2(6.626 \cdot 10^{-34} \mathrm{~J\cdot s})^2}{8(1.67 \cdot 10^{-27} \mathrm{~kg})(1.0 \cdot 10^{-10} \mathrm{~m})^2}\) Solving for \(E_1\), we get: \(E_1 \approx 1.53 \cdot 10^{-25} \mathrm{~J}\)

Calculate the energy at level n=2

Similarly, plug in the values for n=2, m, and L into the energy equation and solve for \(E_2\): \(E_2 = \dfrac{2^2(6.626 \cdot 10^{-34} \mathrm{~J\cdot s})^2}{8(1.67 \cdot 10^{-27} \mathrm{~kg})(1.0 \cdot 10^{-10} \mathrm{~m})^2}\) Solving for \(E_2\), we get: \(E_2 \approx 6.12 \cdot 10^{-25} \mathrm{~J}\)

Calculate the energy at level n=3

Finally, plug in the values for n=3, m, and L into the energy equation and solve for \(E_3\): \(E_3 = \dfrac{3^2(6.626 \cdot 10^{-34} \mathrm{~J\cdot s})^2}{8(1.67 \cdot 10^{-27} \mathrm{~kg})(1.0 \cdot 10^{-10} \mathrm{~m})^2}\) Solving for \(E_3\), we get: \(E_3 \approx 1.38 \cdot 10^{-24} \mathrm{~J}\)

Write the final answer

So, the three lowest energies of the wave function of a proton in a box of width \(1.0 \cdot 10^{-10} \mathrm{~m}\) are: \(E_1 \approx 1.53 \cdot 10^{-25} \mathrm{~J}\), \(E_2 \approx 6.12 \cdot 10^{-25} \mathrm{~J}\), and \(E_3 \approx 1.38 \cdot 10^{-24} \mathrm{~J}\).