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Chapter 29: Problem 20

To make an order-of-magnitude estimate of the energy level spacing's in the nucleus, assume that a nucleon is confined to a one-dimensional box of width \(10 \mathrm{fm}\) (a typical nuclear diameter). Calculate the energy of the ground state.

Short Answer

Expert verified
Answer: The estimated ground state energy for a nucleon confined to a one-dimensional box of width 10 femtometers is approximately 2.04 x 10^(-20) Joules.

Step by step solution

01

Understand the particle in a box model

A particle in a one-dimensional box is a simple quantum mechanics problem where a particle is assumed to be confined within a region of space. The wavefunction of the particle is zero outside the box, and its energy levels are quantized.
02

Know the energy levels equation

The allowed energy levels for a particle confined in a one-dimensional box of width L are given by the equation: \[E_n = \frac{n^2 \hbar^2 \pi^2}{2mL^2}\] where \(E_n\) is the energy corresponding to the quantum number n, \(\hbar\) is the reduced Planck constant, m is the mass of the particle (nucleon in our case), and L is the width of the box.
03

Identify the relevant values

In this problem, we are given the value of L, the width of the box as \(10 \mathrm{fm} = 10^{-14} \mathrm{m}\). The mass of a nucleon (proton or neutron) is approximately \(m \approx 1.67 \times 10^{-27} \mathrm{kg}\). The reduced Planck constant is given by \(\hbar = 1.054 \times 10^{-34} Js\).
04

Calculate the ground state energy

For the ground state, we want to find the lowest energy, which corresponds to the quantum number n=1. Substituting this value, along with the given value of L, the mass of a nucleon, and the reduced Planck constant into the energy levels equation, we can calculate the ground state energy: \[E_1 = \frac{1^2 \times (1.054 \times 10^{-34})^2 \times \pi^2}{2 \times 1.67 \times 10^{-27} \times (10^{-14})^2}\] Solve for \(E_1\): \[E_1 \approx 2.04 \times 10^{-20}J\] The energy of the ground state for a nucleon confined to a one-dimensional box of width \(10 \mathrm{fm}\) is approximately \(2.04 \times 10^{-20} J\).

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