Question

A stationary muon ${\mathit{\mu }}^{\mathbf{-}}$ annihilates with a stationary antimuon$\mathit{\mu }\mathbf{+}$ (same mass, role="math" localid="1657587173645" $\mathbf{1}\mathbf{.}\mathbf{88}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{28}}\mathbf{\hspace{0.33em}}\mathit{k}\mathit{g}$. but opposite charge). The two disappear, replaced by electromagnetic radiation. (a) Why is it not possible for a single photon to result? (b) Suppose two photons result. Describe their possible directions of motion and wavelengths.

Short answer

(a) A single proton carries a particular momentum in all reference frames but the two muons do not carry.

(b) The wavelength of the photons$1.17\times {10}^{-14}\hspace{0.33em}$ is and it is in the opposite direction.

Step-by-step solution

Identification of the given data

The given data is listed below as:

• The mass of the muon ${\mu }^{-}$is role="math" localid="1657587213854" $1.88\times {10}^{-28}\hspace{0.33em}kg$.
• The mass of the muon is $\mu +$is $1.88\times {10}^{-28}\hspace{0.33em}kg$.

Significance of the wavelength

The wavelength is the distance amongst two same points of a particular waveform. The wavelength is the division of the Planck’s constant and the product of the mass of an atom and the velocity of light.

(a) Determination of the for a single photon to result

Light has the equal speed in all reference frames. However, according to the special relativity theory, a particular photon always carries a particular momentum but in the case of muons, in different reference frame, has a zero speed and also zero momentum. Hence, the annihilation of the muons cannot result to a single proton.

Thus, A single proton carries a particular momentum in all reference frames but the two muons does not carry.

(b) Determination of the direction of motion and the wavelength of the photon

The two muons that is stationary initially, will make the photons to move in the opposite direction. Hence, their momentum before and after collision remains equal.

The equation of the wavelength of the photons is expressed as:

$$\begin{gathered} \frac{hc}{\lambda }=m_{\mu }{c}^2 \\ \lambda =\frac{hc}{m_{\mu }{c}^2}\mid \\ \lambda =\frac{h}{2m_{\mu }c} \end{gathered}$$

Here,$\lambda$is the wavelength of the photon,$h$is the Planck’s constant,$c$is the velocity of the light and$m_p$is the mass of the muon.

Substitute $6.626\times {10}^{-34}\mathrm{J}·\mathrm{s}\text{for}h,1.88\times {10}^{-28}\mathrm{kg}\text{for}{m}_p\text{and}3\times {10}^8\mathrm{m}/\mathrm{s}$for$c$in the above equation.

$$\begin{gathered} \lambda =\frac{6.626\times {10}^{-34}\mathrm{kg}·{\mathrm{m}}^2/\mathrm{s}}{\left(1.88\times {10}^{-28}\mathrm{kg}\right)\left(3\times {10}^8\mathrm{m}/\mathrm{s}\right)} \\ =\frac{6.626\times {10}^{-34}\mathrm{kg}·{\mathrm{m}}^2/\mathrm{s}}{\left(5.64\times {10}^{-20}\mathrm{kg}·\mathrm{m}/\mathrm{s}\right)} \\ =1.17\times {10}^{-14}m \end{gathered}$$

Thus, the wavelength of the photons is $1.17\times {10}^{-14}m$and it is in the opposite direction.