Question

A proton with a mass of \(1.673 \cdot 10^{-27} \mathrm{~kg}\) is trapped inside a onedimensional infinite potential well of width \(23.9 \mathrm{nm}\). What is the quantum number, \(n\), of the state that has an energy difference of \(1.08 \cdot 10^{-3} \mathrm{meV}\) with the \(n=2\) state?

Short answer

Answer: The quantum number for the state with the energy difference of 1.08e-3 meV with the n=2 state is n'=3.

Step-by-step solution

Recognize the energy level formula for a one-dimensional infinite potential well

The energy levels for an electron in a one-dimensional infinite potential well of width L are given by the formula: $$ E_n = \frac{n^2 \cdot h^2}{8\cdot m\cdot L^2} $$ Where: - \(E_n\) is the energy level corresponding to the quantum number n, - \(n\) is the quantum number, - \(h = 6.626 \cdot 10^{-34} \mathrm{~J \cdot s}\) is the Planck constant, - \(m\) is the mass of the particle, in this case, the proton with mass \(m = 1.673 \cdot 10^{-27} \mathrm{~kg}\), - \(L\) is the width of the potential well, \(L = 23.9 \mathrm{~nm}\), but we need it in meters.

Convert the width of the potential well to meters

The width of the potential well given is in nanometers. We need to convert it into meters to use it in the energy level formula, so we will multiply by \(10^{-9}\). $$ L = 23.9 \mathrm{~nm} \cdot 10^{-9} \frac{\mathrm{m}}{\mathrm{nm}} = 23.9 \cdot 10^{-9} \mathrm{~m} $$

Calculate the energy difference between the n=2 state and the unknown state

We are given that the energy difference between the n=2 state and this unknown state is \(1.08 \cdot 10^{-3} \mathrm{~meV}\). First, let's convert this energy difference to joules by multiplying it by the electronic charge \(e = 1.6\cdot 10^{-19} \mathrm{~C}\). $$ \Delta E = 1.08 \cdot 10^{-3} \mathrm{~meV} \cdot 1.6 \cdot 10^{-22} \frac{\mathrm{J}}{\mathrm{meV}} = 1.728 \cdot 10^{-25} \mathrm{~J} $$

Write down the energy difference equation

We know that the energy difference between the n=2 state and the state we are looking for is 1.728e-25 J. So we can write the energy difference equation as follows, where \(n'\) is the quantum number for the unknown state, $$ \Delta E = E_{n'} - E_2 = \frac{n'^2 h^2}{8\cdot m\cdot L^2} - \frac{2^2 h^2}{8\cdot m\cdot L^2} = 1.728 \cdot 10^{-25} \mathrm{~J} $$

Rearrange the energy difference equation to find \(n'\)

Rearrange the energy difference equation for calculating \(n'\): $$ n'^2 - 2^2 = \frac{8mL^2\Delta E}{h^2} $$ Now, let's substitute the values of \(m\), \(L\), \(\Delta E\), and \(h\) into the equation, and solve for \(n'^2\): $$ n'^2 - 4 = \frac{8 \cdot 1.673 \cdot 10^{-27} \mathrm{~kg} \cdot (23.9 \cdot 10^{-9} \mathrm{~m})^2 \cdot 1.728 \cdot 10^{-25} \mathrm{~J}}{(6.626 \cdot 10^{-34} \mathrm{~J \cdot s})^2} $$ Solving for \(n'^2\) we get: $$ n'^2 \approx 4.94 $$

Calculate the value of \(n'\) (quantum number)

Since we have an approximate value of \(n'^2\), we can take its square root to find \(n'\). Note that quantum numbers are always whole numbers, and our result for \(n'^2\) is very close to 5. So \(n'\) can be rounded to the nearest whole number. $$ n' \approx \sqrt{4.94} \approx 2.22 $$ The nearest whole number to 2.22 is 2. Since n=1 is the ground state, and we are already considering n=2, we need to pick the next possible value. Therefore, the quantum number for the state with the energy difference of \(1.08\cdot 10^{-3} \mathrm{meV}\) with the \(n=2\) state is \(n'=3\).