Chapter 6: Problem 163

An atom can be formed when a negative muon is captured by a proton. The muon has the same charge as the electron and a mass 207 times that of the electron. Calculate the frequency of the photon emitted when this atom makes the transition from \(n=2\) to the \(n=1\) state. Assume that the muon is orbiting a stationary proton.

Short Answer

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The frequency of the photon emitted when the muon "atom" makes the transition from the \(n=2\) to the \(n=1\) state can be calculated using the formula \(\nu = \dfrac{3 e^4 m_\mu}{32 \epsilon_0^2 h^3}\), where \(m_\mu = 207m_e\). Substitute the values of \(e\), \(m_e\), \(\epsilon_0\), and \(h\) to find the frequency of the photon.

Step by step solution

Identify the given information and necessary formulas

In this exercise, we are given the following information: - Mass of muon is 207 times the mass of electron - Charge of muon is equal to the charge of electron We need the following formulas: - Energy levels for hydrogen-like atoms: \(E_n = -\dfrac{Z^2 e^4 m}{8 \epsilon_0^2 h^2 n^2}\) where \(n\) is the energy level, \(Z\) is the atomic number, \(e\) is the charge of the muon, \(m\) is the mass of the muon, \(\epsilon_0\) is the vacuum permittivity, and \(h\) is the Planck constant. - Energy of a photon: \(E = h \nu\) where \(E\) is the energy of the photon, \(h\) is the Planck constant, and \(\nu\) is the frequency.

Calculate the energy difference between n=2 and n=1 states

First, let's find the energy difference between the n=2 and n=1 states. We will use the energy levels for hydrogen-like atoms formula and substitute the given values. For \( n = 1\) \(E_1 = -\dfrac{(1)^2 e^4 m_\mu}{8 \epsilon_0^2 h^2 (1)^2} = -\dfrac{e^4 m_\mu}{8 \epsilon_0^2 h^2}\) For \(n = 2\) \(E_2 = -\dfrac{(1)^2 e^4 m_\mu}{8 \epsilon_0^2 h^2 (2)^2} = -\dfrac{e^4 m_\mu}{32 \epsilon_0^2 h^2}\) The energy difference, \(\Delta E = E_2 - E_1\), can be calculated as follows: \(\Delta E = -\dfrac{e^4 m_\mu}{32 \epsilon_0^2 h^2} - (-\dfrac{e^4 m_\mu}{8 \epsilon_0^2 h^2}) = \dfrac{3 e^4 m_\mu}{32 \epsilon_0^2 h^2}\)

Calculate the frequency of the photon

Now that we have the energy difference, we can use the energy of a photon formula to find the frequency of the photon. \(\Delta E = h \nu\) We want to find \(\nu\), which is the frequency of the photon: \(\nu = \dfrac{\Delta E}{h} = \dfrac{3 e^4 m_\mu}{32 \epsilon_0^2 h^3}\)

Substituting values and obtain final result

Now that we have the formula to calculate the frequency, we substitute the known values into the equation to obtain the final result: \(m_\mu = 207 m_e\) and \(Z = 1\), for a proton. \(\nu = \dfrac{3 e^4 (207 m_e)}{32 \epsilon_0^2 h^3}\) Now, you can simply substitute the values of \(e\), \(m_e\), \(\epsilon_0\), and \(h\) to find the frequency of the photon emitted during the transition.

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Short Answer

Step by step solution

Identify the given information and necessary formulas

Calculate the energy difference between n=2 and n=1 states

Calculate the frequency of the photon

Substituting values and obtain final result

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