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Chapter 7: Problem 50

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

Short Answer

Expert verified
The ground state energy of a proton confined to a one-dimensional box the size of the uranium nucleus is approximately \(116.52 \mathrm{eV}\).

Step by step solution

01

Convert the radius to the length of the box

Since we have the radius of the uranium nucleus (15.0 fm), we can assume that the one-dimensional box has a length of twice the radius, which represents the diameter of the nucleus. Convert the length from femtometers to meters for calculation purposes: Length (L) = Diameter = 2 × Radius = 2 × 15.0 fm = 30.0 fm = 30.0 × 10^-15 m
02

Calculate the ground state energy using the quantum mechanics formula

The formula to calculate the ground state energy (E) of a particle in a one-dimensional box is given by: E = \(\frac{h^2 n^2}{8mL^2}\) where h is the Planck's constant, n is the quantum number (n=1 for the ground state), m is the mass of the proton, and L is the length of the box. Proton Mass (m) = 1.6726 × 10^-27 kg Planck's Constant (h) = 6.626 × 10^-34 Js Now, plug in the values to calculate the ground state energy: E = \(\frac{(6.626 × 10^{-34} Js)^2 × (1)^2}{8 × (1.6726 × 10^{-27} kg) × (30.0 × 10^{-15} m)^2}\)
03

Convert the energy to electron volts (eV)

Now, we need to convert the energy calculated in joules to electron volts. To do this, we will use the electron volt conversion factor: 1 eV = 1.602 × 10^-19 J So, divide the energy in joules by the conversion factor to get the energy in electron volts: E (eV) = E (J) / (1.602 × 10^-19 J/eV)
04

Calculate the numerical answer

Now, perform the necessary calculations to find the ground state energy of the proton confined in the one-dimensional box: E (J) = \(\frac{(6.626 × 10^{-34})^2}{8 × (1.6726 × 10^{-27}) × (30.0 × 10^{-15})^2} = 1.867 × 10^{-14} J\) E (eV) = 1.867 × 10^-14 J / (1.602 × 10^-19 J/eV) = 116.52 eV Thus, the ground state energy of the proton confined to a one-dimensional box the size of the uranium nucleus is approximately 116.52 eV.

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Most popular questions from this chapter

Assume that an electron in an atom can be treated as if it were confined to a box of width 2.0 A. What is the ground state energy of the electron? Compare your result to the ground state kinetic energy of the hydrogen atom in the Bohr's model of the hydrogen atom.

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.

An electron in a box is in the ground state with energy 2.0 eV. (a) Find the width of the box. (b) How much energy is needed to excite the electron to its first excited state? (c) If the electron makes a transition from an excited state to the ground state with the simultaneous emission of \(30.0-\mathrm{eV}\) photon, find the quantum number of the excited state?

Which one of the following functions, and why, qualifies to be a wave function of a particle that can move along the entire real axis? (a) \(\psi(x)=A e^{-x^{2}}\) (b) \(\psi(x)=A e^{-x} ;\) (c) \(\psi(x)=A \tan x\) (d) \(\psi(x)=A(\sin x) / x ;\) (e) \(\psi(x)=A e^{-|x|}\)

Consider an infinite square well with wall boundaries \(\begin{array}{llllll}x=0 & \text { and } & x=L & \text { Show } & \text { that } & \text { the function }\end{array}\) \(\psi(x)=A \sin k x \quad\) is the solution to the stationary Schrödinger equation for the particle in a box only if \(k=\sqrt{2 m E} / \hbar .\) Explain why this is an acceptable wave function only if \(k\) is an integer multiple of \(\pi / L\)
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