Question

Particle A (with rest energy 200 MeV) is at rest in a lab frame when it decays to particle B (rest energy 100 MeV) and particle C (rest energy 50 MeV). What are the (a) total energy and (b) momentum of B and the (c) total energy and (d) momentum of C?

Short answer

(a) The total energy of particle B is $119\text{MeV}$.

(b) The momentum of particle B is $\frac{64\text{MeV}}{c}$.

(c) The total energy of particle C is $81.3\text{MeV}$.

(d) The momentum of particle C is $\frac{64\text{MeV}}{c}$.

Step-by-step solution

Identification of given data

The total energy of particle A is $E_{tA}=200\text{MeV}$

The rest energy of particle B is $E_B=100\text{MeV}$

The rest energy of particle C is$E_C=50\text{MeV}$

The rest energy of any particle is the energy which is equal to moving particle with speed of light. This relativistic energy varies with the Lorentz factor of the particle.

Determination of Lorentz factors for particle B and particle C

The total energy of particle A is given as:

$$\begin{gathered} E_{tA}=E_{tB}+E_{tC} \\ {E}_{tA}={\gamma }_B{E}_B+{\gamma }_C{E}_C \\ 200\text{MeV}={\gamma }_B\left(100\text{MeV}\right)+{\gamma }_C\left(50\text{MeV}\right) \\ 2{\gamma }_B+{\gamma }_C=4 \end{gathered}$$ …… (1)

Apply the momentum conservation for particle B and C.

$$\begin{gathered} P_B=P_C \\ {m}_B\sqrt{{\gamma }_B^2-1}=m_C\sqrt{{\gamma }_C^2-1} \\ \frac{\sqrt{{\gamma }_B^2-1}}{\sqrt{{\gamma }_C^2-1}}=\frac{m_C{c}^2}{m_B{c}^2} \\ \frac{\sqrt{{\gamma }_B^2-1}}{\sqrt{{\gamma }_C^2-1}}=\frac{E_C}{E_B} \end{gathered}$$

Substitute all the values in the above equation.

$$\begin{gathered} \frac{\sqrt{{\gamma }_B^2-1}}{\sqrt{{\gamma }_C^2-1}}=\frac{50\text{MeV}}{100\text{MeV}} \\ \frac{\sqrt{{\gamma }_B^2-1}}{\sqrt{{\gamma }_C^2-1}}=\frac{1}{2} \\ 4\left({\gamma }_B^2-1\right)={\gamma }_C^2-1 \\ \left(2{\gamma }_B+{\gamma }_C\right)\left(2{\gamma }_B-{\gamma }_C\right)=3 \end{gathered}$$

Substitute value from equation (1) in above expression.

$$\begin{gathered} \left(4\right)\left(2{\gamma }_B-{\gamma }_C\right)=3 \\ 2{\gamma }_B-{\gamma }_C=\frac{3}{4} \end{gathered}$$ …… (2)

Solve equation (1) and equation (3) to find Lorentz factors of particles B and C.

${\gamma }_B=\frac{19}{16}$, ${\gamma }_C=\frac{13}{8}$

Determination of total energy for particle B (a) 

The total energy of particle B is given as

$E_{tB}={\gamma }_B{E}_B$

Substitute all the values in the above equation.

$$\begin{gathered} E_{tB}=\left(\frac{19}{16}\right)\left(100\text{MeV}\right) \\ \approx 119\text{MeV} \end{gathered}$$

Therefore, the total energy of particle B is $119\text{MeV}$.

Determination of momentum of particle B (b)

The momentum of particle B is given as:

$P_B=\left(\frac{E_B}{c}\sqrt{{\gamma }_B^2-1}\right)$

Here, $c$ is the speed of light and its value is $3\times {10}^8\text{m}/\text{s}$

Substitute all the values in the above equation.

$$\begin{gathered} P_B=\left(\frac{100\text{MeV}}{c}\right)\left(\sqrt{{\left(\frac{19}{16}\right)}^2-1}\right) \\ =\frac{64\text{MeV}}{c} \end{gathered}$$

Therefore, the momentum of particle B is $\frac{64\text{MeV}}{c}$.

Determination of total energy for particle C (c)

The total energy of particle C is given as

$E_{tC}={\gamma }_C{E}_C$

Substitute all the values in the above equation.

$$\begin{gathered} E_{tC}=\left(\frac{13}{8}\right)\left(50\text{MeV}\right) \\ =81.3\text{MeV} \end{gathered}$$

Therefore, the total energy of particle C is $81.3\text{MeV}$.

Determination of momentum of particle C (d)

The momentum of particle C is given as:

$P_C=\left(\frac{E_C}{c}\sqrt{{\gamma }_C^2-1}\right)$

Substitute all the values in the above equation.

$$\begin{gathered} P_C=\left(\frac{50\text{MeV}}{c}\right)\left(\sqrt{{\left(\frac{13}{8}\right)}^2-1}\right) \\ =\frac{64\text{MeV}}{c} \end{gathered}$$

Therefore, the momentum of particle C is $\frac{64\text{MeV}}{c}$.