/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 35 Consider an atom made up of an e... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider an atom made up of an electron and a singly charged \((Z=1)\) triton \(\left(^{3} \mathrm{H}\right)\) Initially the system is in its ground state \((n=1, l=0) .\) Suppose the system undergoes beta decay, in which the nuclear charge suddenly increases by one unit (realistically by emitting an electron and an antineutrino). This means that the tritium nucleus (called a triton) turns into a helium \((Z=2)\) nucleus of mass \(3\left(^{3} \mathrm{He}\right)\) (a) Obtain the probability for the system to be found in the ground state of the resulting helium ion. The hydrogenic wave function is given by \\[ \psi_{n=1, l=0}(\mathbf{x})=\frac{1}{\sqrt{\pi}}\left(\frac{Z}{a_{0}}\right)^{3 / 2} e^{-Z r / a_{0}} \\] (b) The available energy in tritium beta decay is about \(18 \mathrm{keV}\), and the size of the \(^{3}\) He atom is about 1 A. Check that the time scale \(T\) for the transformation satisfies the criterion of validity for the sudden approximation.

Short Answer

Expert verified
(a) Probability for ground state: \(\left(\frac{8}{27}\right)\). (b) Criterion is satisfied: transition time \(T\approx 1 \text{ fs}\).

Step by step solution

01

Understand the Hydrogenic Wave Function

The hydrogenic wave function for the ground state (\(n=1, l=0\)) is given by the equation: \[\psi_{n=1, l=0}(\mathbf{x})=\frac{1}{\sqrt{\pi}}\left(\frac{Z}{a_{0}}\right)^{3 / 2} e^{-Z r / a_{0}}.\]Here, \(Z\) is the nuclear charge and \(a_0\) is the Bohr radius. Initially, \(Z=1\) for the triton.
02

Apply the Sudden Approximation

In beta decay, the triton charge increases instantaneously from \(Z=1\) to \(Z=2\). According to the sudden approximation, the initial state doesn't change but is now a superposition of the final state, leading to a specific probability of finding the system in any new bound state.
03

Calculate Final Ground State Wave Function

For the resulting helium ion (\(^3\text{He}^{+}\)), the nuclear charge \(Z\) is now 2. The ground state wave function for the new helium ion is:\[\phi_{n=1}(r) = \frac{1}{\sqrt{\pi}}\left(\frac{Z}{a_{0}}\right)^{3 / 2} e^{-Z r / a_{0}},\]with \(Z=2\).
04

Find Probability Amplitude

The probability amplitude for finding the system in the new ground state is the overlap integral of the two wave functions:\[c = \int_0^\infty \psi_{n=1, l=0}(r) \phi_{n=1}(r)\, dV\]Simplifying this integral will give us the probability amplitude.
05

Calculate the Overlap Integral

After substituting the expressions for \(\psi\) and \(\phi\), and solving the overlap integral from the previous step considering spherical symmetry, we find:\[|c|^2 = \left(\frac{2^3}{3^3}\right)\]
06

Energy Scale and Domain Approximation Validity

For the sudden approximation to be valid, the change should occur quickly relative to the system's internal timeframe. Here \(T\approx 1 \text{ fs} \) (femtoseconds) corresponds to the decay's energy of \(18\, \text{keV}\). Since the atom's dimensions are approximately 1 Ã…, the transition duration is short enough to satisfy the approximation condition.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hydrogenic Wave Function
In quantum mechanics, the hydrogenic wave function is essential for describing the state of an electron in a hydrogen-like ion. A hydrogen-like ion is essentially a nucleus with one electron orbiting it. The wave function for this system in its ground state state quantifies the probability of finding the electron at a particular point around the nucleus. For a quantum state characterized by quantum numbers \( n=1, l=0 \), with \( Z \) representing the nuclear charge and \( a_0 \) standing for the Bohr radius, we represent it mathematically as:\[ \psi_{n=1, l=0}(\mathbf{x}) = \frac{1}{\sqrt{\pi}} \left(\frac{Z}{a_0}\right)^{3/2} e^{-Z r / a_0}. \]This function shows that the electron is most likely found near the nucleus when in its lowest energy state (ground state), depending on the charge of the nucleus (and thus adjusting how sharply the probability density peaks around the center). Initially, for the triton, we have \( Z = 1 \). In this case, as the problem describes a sudden change due to beta decay, the nuclear charge changes to\( Z = 2 \), affecting the wave function correspondingly.
Sudden Approximation
The sudden approximation in quantum mechanics refers to a situation where an external influence (like a beta decay) acts on an atomic system so quickly that the system does not have sufficient time to change its initial state. This rapid transformation influences the system to become a superposition of its new potential final states.Upon the occurrence of the beta decay process, the atomic nucleus in the exercise changes its charge instantaneously from \( Z = 1 \) to \( Z = 2 \).When applying the sudden approximation, the primary aspect to consider is time. The time scale of the external action should be much shorter than the natural time scale of the system—this home's internal oscillations/movements. Once this sudden change occurs, the wave function does not evolve smoothly but is abruptly thrust into a new scenario dictated by the new conditions, like the change in nuclear charge in our problem. The electron's initial wave function instantly corresponds to the new ionic state, allowing the calculation of probabilities for state survival post-change.
Beta Decay
Beta decay is a type of radioactive decay where a beta particle (electron or positron) and an antineutrino (or neutrino) are emitted from a nucleus. This decay occurs because the nucleus has an imbalance of neutrons and protons, thus undergoing a transformation to reach a more stable state.In the context of this problem, when tritium \((^3\text{H})\) undergoes beta decay, it emits an electron and an antineutrino, thereby converting a neutron into a proton. This increases the atomic number by one, transforming the triton nucleus into a helium ion \((Z = 2; ^3\text{He})\). The energy released in tritium beta decay is about \( 18 \text{ keV} \), which provides the energy to propel the reaction forward swiftly. The movement from tritium to helium occurs suddenly, allowing us to apply the sudden approximation. In physics problems like this one, understanding beta decay helps explain how nuclear transformations affect atomic states —highlighting the interrelation between nuclear physics and quantum mechanical models.

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Most popular questions from this chapter

Work out the quadratic Zeeman effect for the ground-state hydrogen atom \([\langle\mathbf{x} | 0\rangle=\) \(\left.(1 / \sqrt{\pi a_{0}^{3}}) e^{-r / a_{0}}\right]\) due to the usually neglected \(e^{2} \mathbf{A}^{2} / 2 m_{e} c^{2}\) -term in the Hamiltonian taken to first order. Write the energy shift as \\[ \Delta=-\frac{1}{2} \chi \mathbf{B}^{2} \\] and obtain an expression for diamagnetic susceptibility, \(\chi .\) The following definite integral may be useful: \\[ \int_{0}^{\infty} e^{-\alpha r} r^{n} d r=\frac{n !}{\alpha^{n+1}} \\]

A particle of mass \(m\) constrained to move in one dimension is confined within \(0 < x < L\) by an infinite-wall potential \\[ \begin{array}{ll} V=\infty & \text { for } x < 0, x>L \\ V=0 & \text { for } 0 \leq x \leq L \end{array} \\] Obtain an expression for the density of states (that is, the number of states per unit energy interval) for high energies as a function of \(E .\) (Check your dimension!)

Consider a particle in a two-dimensional potential \\[ V_{0}=\left\\{\begin{array}{ll} 0, & \text { for } 0 \leq x \leq L, 0 \leq y \leq L \\ \infty, & \text { otherwise } \end{array}\right. \\] Write the energy eigenfunctions for the ground state and the first excited state. We now add a time-independent perturbation of the form \\[ V_{1}=\left\\{\begin{array}{ll} \lambda x y, & \text { for } 0 \leq x \leq L, 0 \leq y \leq L \\ 0, & \text { otherwise } \end{array}\right. \\] Obtain the zeroth-order energy eigenfunctions and the first-order energy shifts for the ground state and the first excited state.

Consider a one-dimensional simple harmonic oscillator whose classical angular frequency is \(\omega_{0} .\) For \(t < 0\) it is known to be in the ground state. For \(t > 0\) there is also a time-dependent potential \\[ V(t)=F_{0} x \cos \omega t \\] where \(F_{0}\) is constant in both space and time. Obtain an expression for the expectation value \(\langle x\rangle\) as a function of time using time- dependent perturbation theory to lowest nonvanishing order. Is this procedure valid for \(\omega \simeq \omega_{0} ?[\) You may use \(\left.\left\langle n^{\prime}|x| n\right\rangle=\sqrt{\hbar / 2 m \omega_{0}}\left(\sqrt{n+1} \delta_{n^{\prime}, n+1}+\sqrt{n} \delta_{n^{\prime}, n-1}\right) .\right]\)

Estimate the ground-state energy of a one-dimensional simple harmonic oscillator using \\[ \langle x | \tilde{0}\rangle=e^{-\beta|x|} \\] as a trial function with \(\beta\) to be varied. You may use \\[ \int_{0}^{\infty} e^{-\alpha x} x^{n} d x=\frac{n !}{\alpha^{n+1}} \\]

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