Problem 59 - University Physics with Modern Physics

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?

Step-by-Step Solution

Answer

(a) 186\(m_e\), (b) -2529.6 eV, (c) 0.653 nm

Step 1: Calculate the Reduced Mass

The reduced mass \( \mu \) of the muonic hydrogen atom is given by the formula: \[ \mu = \frac{m_p \cdot m_\mu}{m_p + m_\mu} \] where \( m_p \) is the mass of the proton and \( m_\mu \) is the mass of the muon. Given that the muon has a mass 207 times that of the electron \( m_e \), we have \( m_\mu = 207m_e \). The mass of the proton is \( m_p \approx 1836m_e \). Substituting these values, \[ \mu = \frac{1836m_e \times 207m_e}{1836m_e + 207m_e} = \frac{380412m_e^2}{2043m_e} \approx 186m_e. \]

Step 2: Calculate Ground Level Energy

The ground level energy \( E_1 \) of a hydrogen-like atom is given by:\[ E_1 = -\frac{\mu e^4}{2\hbar^2 (4\pi\varepsilon_0)^2} \] This is similar to the energy of the hydrogen atom scaled by the reduced mass ratio:\[ E_1 = -13.6 \text{ eV} \times \frac{\mu}{m_e} = -13.6 \text{ eV} \times \frac{186}{1} = -2529.6 \text{ eV}. \]

Step 3: Calculate Wavelength of Radiation Emitted

The energy of the transition from \( n=2 \) to \( n=1 \) level is:\[ \Delta E = E_1 \left(1 - \frac{1}{n^2}\right) = 2529.6 \text{ eV} \left(1 - \frac{1}{4}\right) = 2529.6 \text{ eV} \times \frac{3}{4} = 1897.2 \text{ eV} \]Using the energy-wavelength relation, \( \Delta E = \frac{hc}{\lambda} \), we solve for \( \lambda \):\[ \lambda = \frac{hc}{\Delta E} = \frac{1240 \text{ eV}\cdot\text{nm}}{1897.2 \text{ eV}} \approx 0.653 \text{ nm}. \]

Key Concepts

Reduced Mass Calculation

In physics, when dealing with a system of two particles, such as a muonic hydrogen atom, the concept of reduced mass is crucial for simplifying calculations. It effectively reduces the two-body problem into a simpler one-body problem. For a muonic hydrogen atom, we determine the reduced mass \( \mu \) using the formula:\[\mu = \frac{m_p \cdot m_\mu}{m_p + m_\mu}\]Here, \( m_p \) is the mass of the proton, and \( m_\mu \) is the mass of the muon. The muon is significantly heavier than an electron—specifically, 207 times more massive. Knowing the proton mass \( m_p \approx 1836m_e \) and substituting the values, the reduced mass \( \mu \approx 186m_e \). This is much closer to the mass of the proton than the electron, meaning the muon actually orbits very close to the proton compared to an electron in a normal hydrogen atom.

Ground Level Energy

The ground level energy of hydrogen-like atoms depends on their reduced mass. For a muonic hydrogen atom, the ground level energy \( E_1 \) is given by a variation of the familiar formula for hydrogen:\[E_1 = -\frac{\mu e^4}{2\hbar^2 (4\pi\varepsilon_0)^2}\]Where the constants \( e \), \( \hbar \), and \( \varepsilon_0 \) are the elementary charge, reduced Planck's constant, and permittivity of free space, respectively. To simplify, this energy is essentially scaled from the hydrogen atom's energy \(-13.6 \text{ eV}\) by the ratio of the reduced mass to the electron mass. Thus, for a muonic hydrogen:\[E_1 = -13.6 \text{ eV} \times \frac{186}{1} = -2529.6 \text{ eV}\]The high energy value reflects the powerful electromagnetic binding of the muon close to the proton.

Energy Transition Calculation

In quantum systems like hydrogen atoms, electrons (or in this case, muons) transition between energy levels. The energy difference between these levels involves emission or absorption of radiation. For a muonic hydrogen atom transitioning from \( n=2 \) to \( n=1 \), the energy difference \( \Delta E \) can be calculated as:\[\Delta E = E_1 \left(1 - \frac{1}{n^2}\right)\]Substitute our ground state energy:\[\Delta E = 2529.6 \text{ eV} \left(1 - \frac{1}{4}\right) = 1897.2 \text{ eV}\]This large energy difference compared to a normal hydrogen atom, again, is due to the muon's heavier mass and higher attraction to the proton.

Radiation Wavelength

The wavelength of radiation emitted as a muonic hydrogen atom transitions between energy levels is an essential part of understanding atomic emission spectra. This wavelength \( \lambda \) can be determined by linking it to the energy transition \( \Delta E \) using:\[\Delta E = \frac{hc}{\lambda}\]Where \( h \) is Planck's constant, and \( c \) is the speed of light. To find \( \lambda \), rearrange to get:\[\lambda = \frac{hc}{\Delta E}\]Applying the values for \( hc = 1240 \text{ eV}\cdot\text{nm} \) and \( \Delta E = 1897.2 \text{ eV} \), the wavelength is:\[\lambda \approx \frac{1240 \text{ eV}\cdot\text{nm}}{1897.2 \text{ eV}} \approx 0.653 \text{ nm}\]This extremely short wavelength indicates the radiation is in the X-ray part of the spectrum, which is typical for a system with such high energy transitions like muonic atoms.