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Consider a muonic hydrogen atom, in which an electron is replaced by a muon of mass \(105.66 \mathrm{MeV} / \mathrm{c}^{2}\) that orbits a proton. What are the first three energy levels of the muon in this type of atom?

Short Answer

Expert verified
Answer: To find the first three energy levels of a muonic hydrogen atom, we need to use its reduced mass \(\mu\) and the energy level formula: \(E_n = -\frac{e^4 \mu}{2(4\pi\varepsilon_0)^2n^2h^2}\) By substituting the values of the reduced mass \(\mu\), the charge of the electron \(e\), the vacuum permittivity \(\varepsilon_0\), and the Planck constant \(h\) into the formula, we can calculate the energy levels for \(n = 1, 2, 3\). The numerical values for the first three energy levels of the muonic hydrogen atom can then be determined.

Step by step solution

01

Calculate the reduced mass of the muon-proton system

To find the reduced mass of the muon-proton system, we will use the formula: \(\mu = \frac{m_\mu m_p}{m_\mu + m_p}\) where \(m_\mu\) is the mass of the muon and \(m_p\) is the mass of the proton. The mass numbers given in the question are in \(\mathrm{MeV}/\mathrm{c}^2\), but we need them in kilograms to use them in our energy level formula. To convert the mass numbers to kilograms, we use the following relation: \(mass \; (kg) = \frac{mass \; (\mathrm{MeV}/\mathrm{c}^2)}{c^2} \cdot \frac{1.602 \times 10^{-13} \; \mathrm{J}}{1 \; \mathrm{MeV}}\) where \(c\) is the speed of light in vacuum. For the muon's mass (given as \(105.66 \; \mathrm{MeV}/\mathrm{c}^2\)) in kilograms, we have: \(m_\mu (kg) = \frac{105.66 \; \mathrm{MeV}/\mathrm{c}^2}{c^2} \cdot \frac{1.602 \times 10^{-13} \; \mathrm{J}}{1 \; \mathrm{MeV}}\) The proton's mass is approximately \(938.27 \; \mathrm{MeV}/\mathrm{c}^2\), so its mass in kilograms is: \(m_p (kg) = \frac{938.27 \; \mathrm{MeV}/\mathrm{c}^2}{c^2} \cdot \frac{1.602 \times 10^{-13} \; \mathrm{J}}{1 \; \mathrm{MeV}}\) Now we can calculate the reduced mass of the muon-proton system: \(\mu = \frac{m_\mu m_p}{m_\mu + m_p}\)
02

Substitute the reduced mass into the energy level formula

Now we can substitute the reduced mass \(\mu\) into the energy level formula for the hydrogen-like atom: \(E_n = -\frac{Z^2 e^4 \mu}{2(4\pi\varepsilon_0)^2n^2h^2}\) In this case, \(Z = 1\) as it is a hydrogen-like atom. Therefore, the formula simplifies to: \(E_n = -\frac{e^4 \mu}{2(4\pi\varepsilon_0)^2n^2h^2}\) We can now calculate the energy levels for \(n = 1, 2, 3\).
03

Calculate the first three energy levels

To find the first three energy levels of the muonic hydrogen atom, we can plug in the values for \(n = 1, 2, 3\) into the energy level formula: \(E_1 = -\frac{e^4 \mu}{2(4\pi\varepsilon_0)^2(1)^2h^2}\) \(E_2 = -\frac{e^4 \mu}{2(4\pi\varepsilon_0)^2(2)^2h^2}\) \(E_3 = -\frac{e^4 \mu}{2(4\pi\varepsilon_0)^2(3)^2h^2}\) By plugging in the known values for \(e\), \(\varepsilon_0\), and \(h\), and the reduced mass \(\mu\) calculated in step 1, we can find the numerical values for the first three energy levels of the muonic hydrogen atom.

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Key Concepts

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

Reduced Mass Calculation
Understanding the reduced mass is vital when studying systems with two bodies, such as the muonic hydrogen atom—a variant of a hydrogen atom where a muon substitutes the electron. The reduced mass, \(\mu\), is a critical concept in classical and quantum mechanics since it simplifies the two-body problem into an equivalent one-body problem.

In quantum mechanics, reduced mass allows us to solve the Schrödinger equation for the electron (or muon) in a hydrogen-like atom as though it were moving around a fixed nucleus, despite both particles actually orbiting their common center of mass. This substitution is remarkably useful as it accounts for the influence of both masses on the system's behavior, without overly complicating mathematical treatments.

To calculate the reduced mass, we use the formula: \(\mu = \frac{m_\mu m_p}{m_\mu + m_p}\), where \(m_\mu\) is the mass of the muon and \(m_p\) is the mass of the proton. The masses must be in consistent units, and often in physics problems, we convert them to kilograms using the equivalence \(1 \text{ MeV/c}^2 = 1.602 \times 10^{-13} \text{ J}\).

An accurate calculation of the reduced mass is imperative as it directly influences the energy levels of the system, which are critical to predict and interpret spectral lines and understand atomic interactions.
Energy Levels
In the quantum world of atoms and particles, energy is quantized, meaning objects like electrons in atoms can only inhabit distinct energy levels. The concept of energy levels is analogous to rungs of a ladder that an electron can occupy around the nucleus of an atom.

These energy levels in a muonic hydrogen atom are determined based on the reduced mass of the muon-proton system and are given by a modified version of the Bohr's model for the hydrogen atom. The energy level formula for a hydrogen-like atom is \(E_n = -\frac{Z^2 e^4 \mu}{2(4\pi\varepsilon_0)^2n^2h^2}\), where \(Z\) is the atomic number (which is 1 for hydrogen-like atoms), \(e\) represents the elementary charge, \(\mu\) is the reduced mass of the system, \(\varepsilon_0\) is the vacuum permittivity, \(n\) is the principal quantum number denoting the energy level, and \(h\) is Planck's constant.

The negative sign in the formula indicates that the energy is lower than that of a free particle, and we see that as the quantum number \(n\) increases, the energy levels become less negative (closer to zero), showing that the electron is less tightly bound to the nucleus. For muonic hydrogen, the energy levels are critical for understanding the muon's behavior and its interaction with the proton, affecting how muonic hydrogen emits or absorbs photons.
Quantum Mechanics
Quantum mechanics is the fundamental theory in physics that provides a description of the properties and behavior of matter and energy on the atomic and subatomic levels. It departs from classical mechanics by asserting that energy, momentum, and other quantities are often restricted to discrete values (quantization), objects have characteristics of both particles and waves (wave-particle duality), and there are limits to how accurately the value of physical quantities can be predicted (the uncertainty principle).

One of the central features of quantum mechanics is the wave function, which provides probabilities of finding a particle at a particular location instead of definite predictions. This probabilistic nature is starkly different from the predictable paths of classical physics. In the case of the muonic hydrogen atom, quantum mechanics describes the muon's probable locations as standing waves — the so-called 'orbitals' that correspond to energy levels.

It is through quantum mechanics that we understand the unique behavior of a muon replacing an electron in a hydrogen atom, altering the energy levels and spectral lines. The energy levels are deduced by solving the Schrödinger equation with the appropriate boundary conditions, a primal tool to deciphering the nuances of atomic scale interactions.

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Most popular questions from this chapter

What is the wavelength of the first visible line in the spectrum of doubly ionized lithium? Begin by writing down the formula for the energy levels of the electron in doubly ionized lithium - then consider energy-level differences that give energies in the appropriate (visible) range. Express the answer in terms of "the transition from state \(n\) to state \(n^{\prime}\) is the first visible, with wavelength \(X\)."

A hydrogen atom is in its fifth excited state, with principal quantum number \(n=6 .\) The atom emits a photon with a wavelength of \(410 \mathrm{nm}\). Determine the maximum possible orbital angular momentum of the electron after emission.

Given that the hydrogen atom has an infinite number of energy levels, why can't a hydrogen atom in the ground state absorb all possible wavelengths of light?

In a hydrogen atom, the electron is in the \(n=5\) state. Which of the following sets could correspond to the \(\ell, m\) states of the electron? a) 5,-3 b) 4,-5 c) 3,-2 d) 4,-6

An electron in a hydrogen atom is in the \(2 s\) state. Calculate the probability of finding the electron within a Bohr radius \(\left(a_{0}=0.05295 \mathrm{nm}\right)\) of the proton. The ground-state wave function for hydrogen is: $$ \psi_{2 s}(r)=\frac{1}{4 \sqrt{2 \pi a_{0}^{3}}}\left(2-\frac{r}{a_{0}}\right) e^{-r / 2 a_{0}}. $$ The integral is a bit tedious, so you may want consider using mathematical programs such as Mathcad, Mathematica, etc., or doing the integral online at http://integrals.wolfram.com/index.jsp.

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