Question

Creating a Particle. Two protons (each with rest mass$\mathbf{M}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{67}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{27}}\mathbf{}\mathbf{kg}$ ) are initially moving with equal speeds in opposite directions. The protons continue to exist after a collision that also produces an${\mathit{n}}^{\mathbf{0}}$ particle (see Chapter 44). The rest mass of the${\mathit{n}}^{\mathbf{0}}$ is $\mathbf{m}\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{75}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{28}}\mathbf{}\mathbf{kg}$. (a) If the two protons and the${\mathit{n}}^{\mathbf{0}}$ are all at rest after the collision, find the initial speed of the protons, expressed as a fraction of the speed of light. (b) What is the kinetic energy of each proton? Express your answer in . (c) What is the rest energy of the ${\mathbf{n}}^{\mathbf{0}}$, expressed in ? (d) Discuss the relationship between the answers to parts (b) and (c).

Short answer

a) If the two protons and $n^0$are all at rest after the collision, the initial speed of protons, as a fraction of the speed of light is 0.6331c.

b) The kinetic energy of each proton is 274 MeV.

c) The rest energy of$n^0$ is 548Mev .

d) The sum of the initial kinetic energies of the protons is equal to the rest energy of $n^0$.

Step-by-step solution

Define the kinetic energy and the rest energy

Kinetic energy is the difference between total energy and rest energy.

The total energyE of a particle is

$$\begin{gathered} E=K+m{c}^2 \\ =\frac{mc}{\sqrt{1-v^2/c^2}} \\ =\gamma m{c}^2 \end{gathered}$$

Where,K is the relativistic kinetic energy. The expression ofK is:

$K=\frac{mc}{\sqrt{1-v^2/c^2}}-m{c}^2$

$\gamma$is the Lorentz factor. The expression of$\gamma$ is:

$\gamma =\frac{1}{\sqrt{1-v^2/c^2}}$

The energy$\mathit{m}{\mathit{c}}^{\mathbf{2}}$ associated with rest massm rather than motion is known as rest energy of the particle.

Find the initial speed of protons.

According to the conservation of mass-energy, the sum of the kinetic energies of the protons before collision is equal to the total energies of the$n_0$ particle and the two protons after collision.

So,$2M{c}^2+mc=2\left(\gamma M{c}^2\right)$

$$\begin{gathered} \gamma =1+\frac{m}{2M} \\ =1+\frac{9.75\times {10}^{-28}}{1.67\times {10}^{-27}} \\ =1.292 \end{gathered}$$

So, initial speed of proton is:

$$\begin{gathered} \frac{1}{\sqrt{1-v^2/c^2}}=\frac{1}{1.292} \\ \frac{v}{c}=\sqrt{1-\frac{1}{{\left(1.292\right)}^2}} \\ v=0.6331c \end{gathered}$$

Hence the initial speed of protons is 0.6331c.

Determine the kinetic energy.

$$\begin{gathered} K=\left(\gamma -1\right)M{c}^2 \\ =\left(1.292-1\right)\left(1.67\times {10}^{-27}\right)\left(3.0\times {10}^8\right)\left(\frac{1}{1.602\times {10}^{-13}}\right) \\ =274MeV \end{gathered}$$The kinetic energy is:

Hence, kinetic energy of each proton is 274 MeV.

Determine the rest energy.

The rest energy of$n^0$ is:

$$\begin{gathered} E=m{c}^2 \\ =\left(9.75\times {10}^{-28}\right)\left(3.0\times {10}^8\right)\left(\frac{1}{1.602\times {10}^{-13}}\right) \\ =548\mathrm{MeV} \end{gathered}$$

Hence, the rest energy ofrole="math" localid="1664107189723" $n^0$ is 548 MeV and the sum of the initial kinetic energies of the protons is equal to the rest energy of $n^0$.