Among the elementary subatomic particles of physics is the muon, which decays within a few nanoseconds after formation. The muon has a rest mass 206.8 times that of an electron. Calculate the de Broglie wavelength associated with a muon traveling at a velocity of \(8.85 \times 10^{5} \mathrm{~cm} / \mathrm{s}\)

First, we need to determine the actual mass of the muon. We're given the information that its rest mass is 206.8 times the mass of an electron. We can find the mass of the muon (m) by multiplying this factor with the mass of an electron (me), which is approximately \( 9.11 \times 10^{-28} \mathrm{g} \): \[ m_\mu = 206.8 \times m_e = 206.8 \times (9.11 \times 10^{-28} \mathrm{g}) \]

Now that we have the mass of the muon, we can calculate its momentum by multiplying its mass with its given velocity: \[ p_\mu = m_\mu \cdot v_\mu = (206.8 \times (9.11 \times 10^{-28} \mathrm{g})) \times (8.85 \times 10^{5} \mathrm{~cm} / \mathrm{s}) \]

With the momentum calculated, we can now find the de Broglie wavelength using the formula mentioned earlier: \[ \lambda_\mu = \frac{h}{p_\mu} \] where h is the Planck's constant (\( 6.63 \times 10^{-27} \mathrm{~erg} \cdot \mathrm{s}\)) \[ \lambda_\mu = \frac{6.63 \times 10^{-27} \mathrm{~erg} \cdot \mathrm{s}}{ (206.8 \times (9.11 \times 10^{-28} \mathrm{g})) \times (8.85 \times 10^{5} \mathrm{~cm} / \mathrm{s})} \] Now, simply solve for the de Broglie wavelength: \[ \lambda_\mu \approx 2.05 \times 10^{-11} \mathrm{cm} \] So, the de Broglie wavelength associated with a muon traveling at a velocity of \( 8.85 \times 10^{5} \mathrm{~cm} / \mathrm{s} \) is approximately \( 2.05 \times 10^{-11} \mathrm{cm} \).