Chapter 39: Problem 51
The negative muon has a charge equal to that of an electron but a mass that is 207 times as great. Consider a hydrogenlike atom consisting of a proton and a muon. (a) What is the reduced mass of the atom? (b) What is the ground-level energy (in electron volts)? (c) What is the wavelength of the radiation emitted in the transition from the \(n\) = 2 level to the \(n\) = 1 level?
Short Answer
Step by step solution
Calculate the reduced mass
Calculate the ground-level energy
Calculate the energy difference between levels
Calculate the wavelength of emitted radiation
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Reduced Mass
- \( \mu = \frac{m_1 m_2}{m_1 + m_2} \)
For our hydrogen-like atom consisting of a proton and a muon, the proton's mass \( m_p \) is approximately \( 1.67 \times 10^{-27} \text{ kg} \), and the muon's mass \( m_\mu \) is \( 207 \) times the electron's mass, about \( 9.11 \times 10^{-31} \text{ kg} \). By substituting these values, the calculated reduced mass approximates \( 1.64 \times 10^{-28} \text{ kg} \).
Understanding reduced mass helps simplify the dynamics of atomic systems and ensures our calculations remain accurate without overly complex computations.
Ground-Level Energy
- \( E_1 = - \frac{\mu e^4}{8 \varepsilon_0^2 h^2} \)
By substituting these constants into the formula, we determine the ground-level energy for the proton-muon system is approximately \( -2.54 \text{ eV} \). This value indicates the energy required to liberate the muon from the proton's influence is quite small, characteristic of atomic-scale interactions.
Understanding the ground-level energy serves as a foundation for exploring further quantum interactions and energy states within a system.
Energy Transition
For our system of interest, the energy difference when moving from the second energy level \( n = 2 \) to the first level \( n = 1 \) can be described by:
- \( \Delta E = E_1 \left( 1 - \frac{1}{n^2} \right) \)
This energy difference translates to specific wavelengths of light emitted during the transition, calculated using \( \lambda = \frac{hc}{\Delta E} \). For our case, the wavelength is around \( 650 \text{ nm} \). Such transitions form the basis for atomic spectra and play a central role in understanding atomic and molecular structures.