91Ó°ÊÓ

In the given question particle 1 decays into particles 2 and 3, which move off with equal but oppositely directed momenta.

E₁ = 139.6 MeV.

E₂ = 105.7 MeV.

E₃= 0.

Conservation of momentum

According to the law ofconservationofmomentum, if the system is closedthen the total momentum of the system will remain constant.

Conservation of energy

According to the conservation of energy, the total energy of the system remains constant.

${\mathit{K}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}{\mathbf{E}}_{\mathbf{2}}}\left[{\left(E_1-E_2\right)}^2-E_3^2\right]$

Using the conservation of energy,

$$\begin{gathered} Q=K_2+K_3 \\ =E_1-E_2-E_3 \end{gathered}$$

Where E is the rest energy (mc²).

On rearranging the above equation,

Now, here squaring on both sides

Using the conservation of linear momentum

$\left|P_2\left|=\right|P_3\right|$

$∴{\left(P_2\right)}^2={\left(P_3\right)}^2$

$K_2^2+2K_2{E}_2=K_3^2+2K_3{E}_3$ ……. (ii)

Now we subtract equation (i) and equation (ii)

$K_2^2+2K_2{E}_2-2K_2{E}_1+{\left(E_1-E_2\right)}^2-K_2^2+2K_2{E}_2=K_3^2+2K_3{E}_3-K_3^2-2K_3{E}_3$

Then we get,

$2K_2{E}_1+{\left(E_1-E_2\right)}^2=E_3^2$

$2K_2{E}_2={\left(E_1-E_2\right)}^2-E_3^2$

$∴K_2=\frac{1}{2E}\left[{\left(E_1-E_2\right)}^2-K_3^2\right]$

Hence,$K_2=\frac{1}{2E}\left[{\left(E_1-E_2\right)}^2-K_3^2\right]$

Using the above expression of the kinetic energy,

Substitute 139.6 MeV for E₁, 105.7 MeV for E₂, 0 for k₃ and 13.96 MeV for E in the above equation.

Then we get

$K_2=\frac{1}{2E}\left[{\left(139.6-105.7\right)}^2-0\right]$

$\begin{array}{c}{K}_2=\frac{1}{2×13.96}×{\left(33.9\right)}^2\\ =\frac{{\left(33.9\right)}^2}{27.92}\\ =41.1608 MeV\end{array}$

Hence, the kinetic energy of the antimuon is 41.1608 MeV.