Question

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?

Short answer

Answer

(a) The expression for mass of particle is proved.

(b) The right side of equation becomes equal to mass of particle for low speed of particle.

(c) The ratio of mass of particle with mass of electron is 207 .

Step-by-step solution

Identification of given data

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.

Proof for the mass of particle in terms of kinetic energy and momentum

(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.

Determination of work to accelerate the electron from rest

(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.

Determination of ratio of mass of particle with mass of electron

(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 .