Question

An electron with a speed of 0.95c is emitted by a supernova, where c is the speed of light. What is the magnitude of the momentum of this electron?

Short answer

The magnitude of the momentum of the electron is$8.2\times {10}^{-22}\mathrm{kg}.\mathrm{m}/\mathrm{s}$ .

Step-by-step solution

Identification of given data

The given data can be listed below,

• The speed of electron is,v=0.95c
• The speed of light is,$\mathrm{c}=3\times {10}^8\mathrm{m}/\mathrm{s}$

Concept/Significance of relativistic momentum

Momentum is a four-vector system, according to Relativity. A momentum four-vector is (E, px, py, pz). The first ("temporal") component of momentum is energy, followed by the second and third spatial components of momentum.

Determination of the magnitude of the momentum of this electron

The relativistic momentum of the electron is given by,

$\overrightarrow{p}=\gamma m\left|\overrightarrow{v}\right|$ …(i)

Here, m is the mass of the electron whose value is ,${\mathrm{m}}_{\mathrm{e}}=9\times {10}^{-31}\mathrm{kg},\overrightarrow{\mathrm{v}}$is the velocity of the proton and$\gamma$ is the relativistic factor.

The value of relativistic factor is given by,

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

Here, v is the velocity of the proton and c is the speed of light.

Substitute values in the above,

$$\begin{gathered} \gamma =\frac{1}{\sqrt{1-{\left({\displaystyle \frac{0.95c}{c}}\right)}^2}} \\ =3.20 \end{gathered}$$

$$\begin{gathered} \overrightarrow{\mathrm{p}}=\left(3.20\right)\left(9\times {10}^{-31}\mathrm{kg}\right)\left(0.95\mathrm{c}\right) \\ =\left(28.8\right)\left(0.95\times 3\times {10}^8\right)\mathrm{kg}.\mathrm{m}/\mathrm{s} \\ =8.2\times {10}^{-22}\mathrm{kg}.\mathrm{m}/\mathrm{s} \end{gathered}$$

The magnitude of momentum is calculated by substituting all the values in equation is $8.2\times {10}^{-22}\mathrm{kg}.\mathrm{m}/\mathrm{s}$.