Question

It is not very difficult to accelerate an electron to a speed that is $\mathbf{99}\mathbf{\%}$of the speed of light, because the electron has such a very small mass. What is the ratio of the kinetic energy K to the rest energy $\mathit{m}{\mathit{c}}^{\mathbf{2}}$in this case? In the definition of what we mean by kinetic energy K, $\mathit{E}\mathbf{=}\mathit{m}{\mathit{c}}^{\mathbf{2}}\mathbf{+}\mathit{k}$, you must use the full relativistic expression for E, because v/c is not small compared to 1.

Short answer

The ratio of kinetic energy and rest mass energy is $6.09:1$.

Step-by-step solution

Identification of given data

The given data can be listed below,

• Speed of electron is $99\%$of speed of light i.e., $v=0.99c$

Concept/Significance of rest mass.

Rest mass, also known as invariant mass, describes the fascinating behaviour of objects travelling at relativistic speeds. A massless particle possesses rest mass energy, which allows it to move.

Determination of the ratio of the kinetic energy K to the rest energy.

The kinetic energy of the particle is given by,

$K.E=\gamma m{c}^2-m{c}^2$

Here, $m{c}^2$is the rest mass energy of the particle and $\gamma$is the relativistic factor whose value is,

$\gamma =\frac{1}{\sqrt{1-{\left(v/c\right)}^2}}$.

The value of $\gamma$is given by,

$$\begin{gathered} \gamma =\frac{1}{\sqrt{1-{\left(0.99c/c\right)}^2}} \\ =7.09 \end{gathered}$$

Substitute all values in the kinetic energy equation.

$$\begin{gathered} K.E=7.09m{c}^2-m{c}^2 \\ =6.09m{c}^2 \end{gathered}$$

The ratio of kinetic energy with rest mass energy is given by,

$$\begin{gathered} \frac{K.E}{m{c}^2}=\frac{6.09m{c}^2}{m{c}^2} \\ =6.09:1 \end{gathered}$$

Thus, the ratio of kinetic energy and rest mass energy is $6.09:1$.