Question

An electron with a mass of \(9.109 \cdot 10^{-31} \mathrm{~kg}\) is trapped inside a onedimensional infinite potential well of width \(13.5 \mathrm{nm}\). What is the energy difference between the \(n=5\) and the \(n=1\) states?

Short answer

Based on the given information and the calculations above, the energy difference between the n=5 and n=1 states of an electron trapped inside a one-dimensional infinite potential well of width 13.5 nm is approximately \(1.24 \times 10^{-19} \mathrm{J}\).

Step-by-step solution

Convert the width of the potential well to meters

We are given the width of the potential well as 13.5 nm, which we need to convert it to meters. \(1 \mathrm{~nm}\) = \(10^{-9} \mathrm{~m}\) So, the width of the potential well in meters is: \(L = 13.5 \times 10^{-9} \mathrm{~m}\)

Calculate the energy levels for n=1 and n=5

Now, we'll calculate the energy levels for n=1 and n=5 using the formula: \(E_n = \dfrac{n^2 \cdot h^2}{8 \cdot m \cdot L^2}\) \(h = 6.626 \cdot 10^{-34} \mathrm{~Js}\) (Planck's constant) \(m = 9.109 \cdot 10^{-31} \mathrm{~kg}\) (mass of the electron) For \(n=1\): \(E_1 = \dfrac{1^2 \cdot (6.626 \cdot 10^{-34})^2}{8 \cdot (9.109 \cdot 10^{-31}) \cdot (13.5 \times 10^{-9})^2}\) For \(n=5\): \(E_5 = \dfrac{5^2 \cdot (6.626 \cdot 10^{-34})^2}{8 \cdot (9.109 \cdot 10^{-31}) \cdot (13.5 \times 10^{-9})^2}\)

Calculate the energy difference

Subtract the energy of the first state, \(E_1\), from the energy of the fifth state, \(E_5\), to find the energy difference: \(\Delta E = E_5 - E_1\) Now, calculate the energy difference: \(\Delta E = \left( \dfrac{5^2 \cdot (6.626 \cdot 10^{-34})^2}{8 \cdot (9.109 \cdot 10^{-31}) \cdot (13.5 \times 10^{-9})^2} \right) - \left( \dfrac{1^2 \cdot (6.626 \cdot 10^{-34})^2}{8 \cdot (9.109 \cdot 10^{-31}) \cdot (13.5 \times 10^{-9})^2} \right)\) After calculating the expression above, we find out the energy difference between the \(n=5\) and the \(n=1\) states: \(\Delta E \approx 1.24 \times 10^{-19} \mathrm{J}\)