Question:(a) If m is a particle’s mass, p is its momentum magnitude, and K is its kinetic energy, show that

$\mathbf{m}\mathbf{=}\frac{{\left(\mathbf{pc}\right)}^{\mathbf{2}}\mathbf{-}{\mathbf{K}}^{\mathbf{2}}}{\mathbf{2}{\mathbf{Kc}}^{\mathbf{2}}}$

(b) For low particle speeds, show that the right side of the equation reduces to m. (c) If a particle has K = 55.0 MeV when p=121 MeV/c, what is the ratio m/me of its mass to the electron mass?

The kinetic energy of particle is $K=55\text{MeV}$

The momentum of particle is $p=121\text{Mev}/\text{c}$

The mass of the particle is m

The total energy of a particle has two components. One is the rest energy of particle and other is the kinetic energy of the particle due to the motion of particle.

(a)

The rest energy of particle is given as:

$E_0=m{c}^2$

The total energy in terms of momentum and rest energy of the particle is given as:

$E^2={\left(pc\right)}^2+{\left(m{c}^2\right)}^2$

The total energy of particle is given as:

$$\begin{gathered} E=E_0+K \\ E=m{c}^2+K......\left(1\right) \end{gathered}$$

Square both sides of the equation (1).

$$\begin{gathered} E^2=\left(m{c}^2+K\right) \\ {E}^2={\left(m{c}^2\right)}^2+K^2+2Km{c}^2......\left(2\right) \end{gathered}$$

Substitute all the values in equation (2).

$$\begin{gathered} {\left(pc\right)}^2+{\left(m{c}^2\right)}^2={\left(m{c}^2\right)}^2+K^2+2Km{c}^2 \\ {\left(pc\right)}^2=K^2+2Km{c}^2 \\ m=\frac{{\left(pc\right)}^2-K^2}{2K{c}^2}......\left(3\right) \end{gathered}$$

Therefore, the expression for mass of particle is proved.

(b)

The expression for momentum and kinetic energy of particle for low speed are given as:

$$\begin{gathered} p=mv \\ K=\frac{1}{2}m{v}^2 \end{gathered}$$

Substitute these values in equation (3).

$$\begin{gathered} m=\frac{{\left(mvc\right)}^2-{\left(\frac{1}{2}m{v}^2\right)}^2}{2\left(\frac{1}{2}m{v}^2\right)c^2} \\ m=m \end{gathered}$$

Therefore, the right side of equation becomes equal to mass of particle for low speed of particle.

(c)

Substitute the values of kinetic energy and momentum in equation (3)

$$\begin{gathered} m=\frac{{\left[\left(121\text{Mev}/\text{c}\right)c\right]}^2-{\left(55\text{MeV}\right)}^2}{2\left(55\text{MeV}\right)c^2} \\ m=105.6\text{MeV}/c^2 \end{gathered}$$

The mass of electron in electron volt is $0.511\text{MeV}/c^2$ and the ratio of mass of particle with mass of electron is given as:

$$\begin{gathered} \frac{m}{m_e}=\frac{105.6\text{MeV}/c^2}{0.511\text{MeV}/c^2} \\ \frac{m}{m_e}=207 \end{gathered}$$

Therefore, the ratio of mass of particle with mass of electron is 207 .