Question

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

(a) \(1.64 \times 10^{-28} \text{ kg}\) (b) \(-2.54 \text{ eV}\) (c) \(650 \text{ nm}\)

Step-by-step solution

Calculate the reduced mass

The reduced mass \( \mu \) of a two-body system is calculated using the formula: \[ \mu = \frac{m_1 m_2}{m_1 + m_2} \]where \( m_1 \) is the mass of the proton and \( m_2 \) is the mass of the muon. The mass of the proton \( m_p \) is approximately \( 1.67 \times 10^{-27} \text{ kg} \) and the mass of the muon \( m_\mu \) is \( 207 \) times the mass of the electron \( m_e \). With \( m_e \approx 9.11 \times 10^{-31} \text{ kg} \), we find: \[ m_\mu = 207 \times 9.11 \times 10^{-31} = 1.89 \times 10^{-28} \text{ kg} \]Plug the values into the reduced mass formula: \[ \mu = \frac{(1.67 \times 10^{-27})(1.89 \times 10^{-28})}{1.67 \times 10^{-27} + 1.89 \times 10^{-28}} \approx 1.64 \times 10^{-28} \text{ kg} \].

Calculate the ground-level energy

The ground-level energy for a hydrogenlike atom is given by \[ E_1 = - \frac{\mu e^4}{8 \varepsilon_0^2 h^2} \]where \( \mu \) is the reduced mass, \( e \) is the charge of the electron \( (1.6 \times 10^{-19} \text{ C}) \), \( \varepsilon_0 \) is the permittivity of free space \( (8.85 \times 10^{-12} \text{ C}^2/\text{N} \cdot \text{m}^2) \), and \( h \) is Planck's constant \( (6.63 \times 10^{-34} \text{ J} \cdot \text{s}) \). Substituting the known values: \[ E_1 = - \frac{(1.64 \times 10^{-28})(1.6 \times 10^{-19})^4}{8 \times (8.85 \times 10^{-12})^2 \times (6.63 \times 10^{-34})^2} \approx -2.54 \text{ eV} \].

Calculate the energy difference between levels

The energy of a level \( n \) in a hydrogenlike atom is given by \[ E_n = \frac{E_1}{n^2} \]The transition from \( n = 2 \) to \( n = 1 \) involves an energy difference: \[ \Delta E = E_1 \left( 1 - \frac{1}{2^2} \right) = E_1 \left( 1 - \frac{1}{4} \right) = \frac{3}{4}E_1 \]Using \( E_1 = -2.54 \text{ eV} \), we calculate: \[ \Delta E = \frac{3}{4} \times (-2.54) = -1.905 \text{ eV} \].

Calculate the wavelength of emitted radiation

The wavelength \( \lambda \) is calculated using the energy-wavelength relation: \[ \Delta E = \frac{hc}{\lambda} \]Solving for \( \lambda \): \[ \lambda = \frac{hc}{\Delta E} \]Substitute \( h = 6.63 \times 10^{-34} \text{ J} \cdot \text{s} \), \( c = 3 \times 10^8 \text{ m/s} \), and \( \Delta E = 1.905 \times 1.6 \times 10^{-19} \text{ J} \): \[ \lambda = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{1.905 \times 1.6 \times 10^{-19}} = 650 \text{ nm} \].

Key concepts

Reduced Mass

In quantum mechanics, the concept of reduced mass is crucial when dealing with two-body systems like atoms. It helps simplify the equations needed to describe the motion of two interacting particles, such as a proton and a muon. The reduced mass \( \mu \) is calculated using the formula:

• \( \mu = \frac{m_1 m_2}{m_1 + m_2} \)

Here, \( m_1 \) and \( m_2 \) are the masses of the two particles in question.
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

The ground-level energy of a system describes its lowest energy state. In quantum mechanics, for a hydrogen-like atom, this can be calculated using the formula:

• \( E_1 = - \frac{\mu e^4}{8 \varepsilon_0^2 h^2} \)

Here, \( \mu \) represents the reduced mass, \( e \) is the charge of the electron at \( 1.6 \times 10^{-19} \text{ C} \), \( \varepsilon_0 \) is the permittivity of free space, and \( h \) denotes Planck’s constant.
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

In an atom, electrons or similar particles like muons can move between energy levels, resulting in an energy transition. This transition often involves the release or absorption of energy in the form of electromagnetic radiation.
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:

• \( \Delta E = E_1 \left( 1 - \frac{1}{n^2} \right) \)

Instead of calculating each level individually, understand that the transition involves more energy at lower levels. Here, the transition results in an energy difference of approximately \( -1.905 \text{ eV} \).
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.