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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

Consider the energies allowed for bound states of a half-harmonic oscillator, namely, a potential that is $$ U(x)=\left\\{\begin{array}{l} \frac{1}{2} m \omega_{0}^{2} x^{2} \\ \infty \end{array}\right. \text { for }\left\\{\begin{array}{l} x>0 \\ x \leq 0 \end{array}\right. $$ Using simple arguments based on the characteristics of good wave functions, what are the energies allowed for bound states in this potential?

An oxygen molecule has a vibrational mode that behaves approximately like a simple harmonic oscillator with frequency \(2.99 \cdot 10^{14} \mathrm{rad} / \mathrm{s} .\) Calculate the energy of the ground state and the first two excited states.

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

An electron is confined in a one-dimensional infinite potential well of \(1.0 \mathrm{nm}\). Calculate the energy difference between a) the second excited state and the ground state, and b) the wavelength of light emitted by this radiative transition.

The probability of finding an electron in a hydrogen atom is directly proportional to a) its energy. b) its momentum. c) its wave function. d) the square of its wave function. e) the product of the position coordinate and the square of the wave function. f) none of the above.
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