91Ó°ÊÓ

In the study of the muonic hydrogen atom, calculating the reduced mass is vital to understanding its behavior. The reduced mass, denoted as \( \mu \), is crucial in simplifying the analysis of two-body systems like a proton and a muon. You can think of it as an adjusted mass that accounts for both bodies' contributions without focusing on each separately. The formula used is \( \mu = \frac{m_1 m_2}{m_1 + m_2} \), where \( m_1 \) and \( m_2 \) are the masses of the proton and muon, respectively. For this exercise:

• Proton mass \( m_p = 1.6726 \times 10^{-27} \; \text{kg} \)
• Muon mass \( m_\mu = 207 \times 9.1094 \times 10^{-31} \; \text{kg} \)

Substituting these values provides us with a reduced mass \( \mu = 1.69246 \times 10^{-27} \; \text{kg} \). This adjusted mass accounts for the muon's higher mass compared to the electron in a standard hydrogen atom, impacting the atomic characteristics significantly.

The ground-level energy of hydrogen-like atoms is a measure of the energy required to remove the orbiting particle from its ground state, where it's most stable. For our muonic hydrogen atom, we calculate this ground-state energy using the formula: \[ E_n = -\frac{\mu e^4}{2(4\pi \epsilon_0 \hbar)^2} \times \frac{1}{n^2} \]Here are the constants you'll use:

• Elementary charge \( e = 1.602 \times 10^{-19} \; \text{C} \)
• Reduced mass \( \mu = 1.69246 \times 10^{-27} \; \text{kg} \)
• Planck's reduced constant \( \hbar = 1.055 \times 10^{-34} \; \text{J}\cdot\text{s} \)
• Permittivity of vacuum \( \epsilon_0 = 8.854 \times 10^{-12} \; \text{F/m} \)

For \( n = 1 \), the ground-level energy comes out to \( E_1 = -2.55 \; \text{eV} \). This negative sign indicates the bound state of the muon in the atom, meaning energy must be supplied to free it from its current state.

When atoms experience changes in energy levels, they emit or absorb radiation of specific wavelengths. For the hydrogen-like atom with a muon, the transition from the second energy level (\( n=2 \)) to the ground level (\( n=1 \)) emits radiation, which we can calculate by following these steps:- First, find the energy of level \( n=2 \), \( E_2 = -\frac{2.55}{4} = -0.6375 \; \text{eV} \).- Then, determine the energy difference between these levels: \( \Delta E = E_2 - E_1 = -0.6375 - (-2.55) = 1.9125 \; \text{eV} \).To convert this energy difference into a wavelength, we utilize the equation: \[ \lambda = \frac{hc}{\Delta E} \]Where:

• \( h = 6.626 \times 10^{-34} \; \text{J}\cdot\text{s} \)
• \( c = 3 \times 10^8 \; \text{m/s} \)

Substituting these values, we find \( \lambda = 6.57 \times 10^{-9} \; \text{m} \), or 6.57 nanometers. These calculations help us understand the spectral lines emitted by such atoms, contributing to our knowledge of atomic transitions.