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Chapter 7: Problem 51

What is the ground state energy (in eV) of an \(\alpha\) -particle confined to a one-dimensional box the size of the uranium nucleus that has a radius of approximately 15.0 fm?

Short Answer

Expert verified
The ground state energy of the α-particle confined to a one-dimensional box the size of the uranium nucleus is approximately 93.7 eV.

Step by step solution

01

Determine the box length

To calculate the length of the one-dimensional box, we will consider the diameter of the uranium nucleus since it resembles a sphere. The diameter is twice the radius. Using the given radius (15.0 fm), we determine the length in meters: Length (L) = 2 * radius = \(2 \times 15 \text{ fm}\) Convert femtometers (fm) to meters: Length (L) = \(2 \times 15\)e-15 m = 30e-15 m
02

Use the formula for the ground state energy

The formula for the ground state energy of a particle confined in a one-dimensional box is: \[ E_{n} = \dfrac{n^2 h^2}{8mL^2} \] Where, E is the ground state energy, n is the quantum number (we use n = 1 for the ground state), h is the Planck constant (\(6.626 \times 10^{-34} \text{ J·s}\)), m is the mass of the particle, and L is the length of the box. For an α-particle, m = 4 (mass of a proton) x \(1.67 \times 10^{-27} \text{ kg}\) Calculating the mass: m = \(4 \times 1.67 \times 10^{-27} \text{ kg}\) = \(6.68 \times 10^{-27} \text{ kg}\) Now, let's calculate the energy: \[ E_{1} = \dfrac{1^2 (6.626 \times 10^{-34} \text{ J·s})^2}{8(6.68 \times 10^{-27} \text{ kg})(30 \times 10^{-15} \text{ m})^2} \]
03

Convert the energy from Joules to electron volts (eV)

We can calculate the ground state energy in Joules by solving the previous expression. But we need to convert it to electron volts (eV) using the relation, 1 eV = 1.602e-19 J. Let's solve for E: \[ E_{1} (J) = \dfrac{1^2 (6.626 \times 10^{-34} \text{ J·s})^2}{8(6.68 \times 10^{-27} \text{ kg})(30 \times 10^{-15} \text{ m})^2} = 1.50 \times 10^{-14} \text{ J} \] Now, convert to electron volts (eV): \[ E_{1} (eV) = \dfrac{1.50 \times 10^{-14} \text{ J}}{1.602 \times 10^{-19} \text{ J/eV}} \approx 93.7 \text{ eV} \] The ground state energy of the α-particle confined to a one-dimensional box the size of the uranium nucleus is approximately 93.7 eV.

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Most popular questions from this chapter

Consider an infinite square well with wall boundaries \(x=0 \quad\) and \(\quad x=L . \quad\) Explain \(\quad\) why the function \(\psi(x)=A \cos k x \quad\) is not a solution to the stationary Schrödinger equation for the particle in a box.

Is it possible that when we measure the energy of a quantum particle in a box, the measurement may return a smaller value than the ground state energy? What is the highest value of the energy that we can measure for this particle?

A 6.0-eV electron impacts on a barrier with height \(11.0 \mathrm{eV} .\) Find the probability of the electron to tunnel through the barrier if the barrier width is (a) \(0.80 \mathrm{nm}\) and (b) \(0.40 \mathrm{nm}\)

Find the expectation value \(\left\langle x^{2}\right\rangle\) of the square of the position for a quantum harmonic oscillator in the ground state. Note: \(\int_{-\infty}^{+\infty} d x x^{2} e^{-a x^{2}}=\sqrt{\pi}\left(2 a^{3 / 2}\right)^{-1}\)

. Estimate the ground state energy of the quantum harmonic oscillator by Heisenberg's uncertainty principle. Start by assuming that the product of the uncertainties \(\Delta x\) and \(\Delta p\) is at its minimum. Write \(\Delta p\) in terms of \(\Delta x\) and assume that for the ground state \(x \approx \Delta x\) and \(p \approx \Delta p\) then write the ground state energy in terms of \(x .\) Finally, find the value of \(x\) that minimizes the energy and find the minimum of the energy.
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