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Chapter 37: Problem 43

Calculate the ground state energy for an electron confined to a cube with sides equal to twice the Bohr radius \((R=0.0529 \mathrm{nm}) .\) Determine the spring constant that would give this same ground state energy for a harmonic oscillator.

Short Answer

Expert verified
Answer: The spring constant for a harmonic oscillator that would give the same ground state energy as an electron confined to the given cube is approximately \(1.158 \times 10^{-1} \ N/m\).

Step by step solution

01

Calculate the dimensions of the cube

We are given that each side of the cube is twice the Bohr radius, so we need to find the length of the side. The Bohr radius, \(R\), is given as \(0.0529 \ nm\) (nanometers). \(side\_length = 2 * R = 2 * 0.0529 \ nm = 0.1058 \ nm\)
02

Calculate the ground state energy of an electron in a cube using particle-in-a-box model

Using the particle in a box model, the energy levels of an electron can be calculated by the following formula: \(E_n = \frac{n^2 * h^2}{8 * m * L^2}\) where \(n\) is the principal quantum number (for the ground state, \(n = 1\)), \(h\) is Planck's constant (\(6.626 \times 10^{-34} \ Js\)), \(m\) is the mass of the electron (\(9.109 \times 10^{-31} \ kg\)), and \(L\) is the side length of the cube (which we calculated in Step 1). \(E_1 = \frac{1^2 * (6.626 \times 10^{-34})^2}{8 * (9.109 \times 10^{-31}) * (0.1058 \times 10^{-9})^2} = 1.483 \times 10^{-19} \ J\) We now have the ground state energy for the electron confined within the cube.
03

Calculate the ground state energy for a quantum harmonic oscillator

The ground state energy for a quantum harmonic oscillator is given by the following equation: \(E_0 = \frac{1}{2} h \omega\) where \(\omega\) is the angular frequency of the oscillator and is related to the spring constant, \(k\), via the equation: \(\omega = \sqrt{\frac{k}{m}}\) We are given the ground state energy for the confined electron already, so our task is to determine the spring constant that would give us the same ground state energy as the confined electron.
04

Calculate the spring constant for the harmonic oscillator

Knowing that the ground state energies for both systems must be equal, we can set to equal: \(\frac{1}{2} h \omega = E_1\) Now, we can use the relation between \(\omega\) and the spring constant to rewrite the equation in terms of \(k\): \(\frac{1}{2} h \sqrt{\frac{k}{m}} = E_1\) Next, we can solve for the spring constant, \(k\): \(k = \frac{4 * m * E_1^2}{h^2}\) Plugging in the values, we get: \(k = \frac{4 * (9.109 \times 10^{-31} \ kg) * (1.483 \times 10^{-19} \ J)^2}{(6.626 \times 10^{-34} \ Js)^2} = 1.158 \times 10^{-1} \ N/m\) Hence, the required spring constant is approximately \(1.158 \times 10^{-1} \ N/m\).

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Most popular questions from this chapter

The ground state wave function for a harmonic oscillator is given by \(\Psi_{0}(x)=A_{2} e^{-x^{2} / 2 b^{2}}\). a) Determine the normalization constant \(A\). b) Determine the probability that a quantum harmonic oscillator in the \(n=0\) state will be found in the classically forbidden region.

Consider an electron in a three-dimensional box-with infinite potential walls- of dimensions \(1.00 \mathrm{nm} \times 2.00 \mathrm{nm} \times 3.00 \mathrm{nm}\). Find the quantum numbers \(n_{x}, n_{y}, n_{z}\) and energies in \(\mathrm{eV}\) of the six lowest energy levels. Are any of these levels degenerate, that is, do any distinct quantum states have identical energies?

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An approximate one-dimensional quantum well can be formed by surrounding a layer of GaAs with layers of \(\mathrm{Al}_{x} \mathrm{Ga}_{1-x}\) As. The GaAs layers can be fabricated in thicknesses that are integral multiples of the single-layer thickness, \(0.28 \mathrm{nm}\). Some electrons in the GaAs layer behave as if they were trapped in a box. For simplicity, treat the box as an infinite one-dimensional well and ignore the interactions between the electrons and the Ga and As atoms (such interactions are often accounted for by replacing the actual electron mass with an effective electron mass). Calculate the energy of the ground state in this well for these cases: a) 2 GaAs layers b) 5 GaAs layers
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