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Chapter 37: Problem 25

Determine the two lowest energies of a wave function of an electron in a box of width \(2.0 \cdot 10^{-9} \mathrm{~m} .\)

Short Answer

Expert verified
Answer: The two lowest energy levels of the electron in the box are: - Lowest energy level (\(n=1\)): \(E_1 \approx 3.768 \cdot 10^{-19} \mathrm{~J}\) - Second lowest energy level (\(n=2\)): \(E_2 \approx 1.507 \cdot 10^{-18} \mathrm{~J}\)

Step by step solution

01

Write down the given parameters.

The width of the box is given as \(L=2.0 \cdot 10^{-9} \mathrm{~m}\). The mass of the electron is \(m = 9.109 \times 10^{-31} \mathrm{~kg}\), and the Planck constant is \(h = 6.626 \times 10^{-34} \mathrm{~J \cdot s}\).
02

Use the energy formula for n=1.

To find the lowest energy level, we need to calculate the energy for \(n=1\). The formula for the energy level is: $$ E_{n} = \frac{n^2h^2}{8mL^2}. $$ Now, plug in the values for \(n=1\), \(h\), \(m\), and \(L\): $$ E_{1}=\frac{(1)^{2}(6.626\cdot10^{-34}\mathrm{~J\cdot s})^{2}}{8(9.109\cdot10^{-31}\mathrm{~kg})(2.0\cdot10^{-9}\mathrm{~m})^{2}}. $$
03

Calculate the energy for n=1.

Compute the energy for \(n=1\) as follows: $$ E_{1}=\frac{(6.626\cdot10^{-34})^{2}}{8(9.109\cdot10^{-31})(4.0\cdot10^{-18})}\mathrm{~J} \approx 3.768 \cdot 10^{-19} \mathrm{~J}. $$ The lowest energy level of the electron is approximately \(3.768 \cdot 10^{-19} \mathrm{~J}\).
04

Use the energy formula for n=2.

To find the second lowest energy level, calculate the energy for \(n=2\). Use the same energy level formula as in step 2: $$ E_{n} = \frac{n^2h^2}{8mL^2}. $$ Plug in the values for \(n=2\), \(h\), \(m\), and \(L\): $$ E_{2}=\frac{(2)^{2}(6.626\cdot10^{-34}\mathrm{~J\cdot s})^{2}}{8(9.109\cdot10^{-31}\mathrm{~kg})(2.0\cdot10^{-9}\mathrm{~m})^{2}}. $$
05

Calculate the energy for n=2.

Compute the energy for \(n=2\) as follows: $$ E_{2}=\frac{(4)(6.626\cdot10^{-34})^{2}}{8(9.109\cdot10^{-31})(4.0\cdot10^{-18})}\mathrm{~J} \approx 1.507 \cdot 10^{-18} \mathrm{~J}. $$ The second lowest energy level of the electron is approximately \(1.507 \cdot 10^{-18} \mathrm{~J}\).
06

Report the two lowest energies.

The two lowest energy levels of the electron in the box are: - Lowest energy level (\(n=1\)): \(E_1 \approx 3.768 \cdot 10^{-19} \mathrm{~J}\) - Second lowest energy level (\(n=2\)): \(E_2 \approx 1.507 \cdot 10^{-18} \mathrm{~J}\)

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Most popular questions from this chapter

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

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.

Two long, straight wires that lie along the same line have a separation at their tips of \(2.00 \mathrm{nm}\). The potential energy of an electron in the gap is about \(1.00 \mathrm{eV}\) higher than it is in the conduction band of the two wires. Conduction-band electrons have enough energy to contribute to the current flowing in the wire. What is the probability that a conduction electron in one wire will be found in the other wire after arriving at the gap?

37.2 In an infinite square well, for which of the following states will the particle never be found in the exact center of the well? a) the ground state b) the first excited state c) the second excited state d) any of the above e) none of the above

For a particle trapped in an infinite square well of length \(L\), what happens to the probability that the particle is found between 0 and \(L / 2\) as the particle's energy increases?
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