What is the momentum of a proton accelerated through $\mathbf{1}\mathbf{}$gigavolts (GV)?

Relativistic momentum of an object moving at relativistic speed; rest mass of an object measured by an observer at rest relative to the object.

The relativistic relation for momentum is

$p=\gamma mv$

Here, m is the mass and v is the velocity.

The relativistic factor is define by,

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

Now, we have to find velocity (v) gained by the proton accelerating under 1 GV of potential difference.

Relativistic kinetic energy is equal to change in potential energy as the charged particle is accelerated under a potential difference.

$$\begin{gathered} KE=\Delta U \\ KE=e\Delta V \end{gathered}$$ ….. (2)

Here, KE is the kinetic energy, $\Delta U$is the change in potential energy, e is the charge, and $\Delta V$is the potential difference.

Andas the electron is traveling at a very high speed considering relativistic Kinetic energy as below.

$\begin{array}{rcl}KE& =& Totalenergy-m_o{c}^2\\ & =& m{c}^2-m_o{c}^2\\ & =& \gamma m_o{c}^2-m_o{c}^2\\ KE& =& \left(\gamma -1\right)m_o{c}^2\\ & & \end{array}$ ….. (3)

Here, m_o is the rest mass which is equal to the mass of the proton. Therefore, by comparing equation (2) and (3), you get

$\left(\gamma -1\right)m_p{c}^2=e\Delta V$ ….. (4)

Here,

The mass of proton,$m_p=1.67\times {10}^{-27}kg$

The speed of light,$c=3\times {10}^8\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

The charge,$e=1.6\times {10}^{-19}C$

The potential difference,$\Delta V=1\times {10}^{9}V$

Putting the values for mass of the proton, the charge and the potential difference into equation (4), and you have

$\begin{array}{rcl}\left(\gamma -1\right)\left(1.67\times {10}^{-27}kg\right)\left(3\times {10}^8\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\right)& =& \left(1.6\times {10}^{-19}C\right)\left(1\times {10}^{9}V\right)\\ \left(\gamma -1\right)\left(15.03\times {10}^{-11}\right)& =& \left(1.6\times {10}^{-10}\right)\\ \left(\gamma -1\right)& =& 1.06\\ \gamma & =& 2.06\end{array}$

Substitute 2.06 for$\gamma$ into equation (1), and you have

$$\begin{gathered} 2.06=\frac{1}{\sqrt{1-{\displaystyle \raisebox{1ex}{$v^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.}}} \\ 1-\raisebox{1ex}{$v^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.=\frac{1}{4.2436} \\ \raisebox{1ex}{$v^2$}\!\left/ \!\raisebox{-1ex}{$c^2$}\right.=1-0.2356 \\ {v}^2=0.7644c^2 \\ v=0.8743c \end{gathered}$$

Then, using Lorentz factor, velocity of proton will simply be v = 0.8743c,.

Now, as all the parameter needed for relativistic momentum are determined, you obtain,

role="math" localid="1659086536584" $\begin{array}{rcl}p& =& \gamma mv\\ & =& \left(2.06\right)\left(1.67\times {10}^{-27}kg\right)\left(0.8743\times 3\times {10}^8\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\right)\\ & =& 9.023\times {10}^{-19}\text{kg.}\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.\end{array}$

Hence, the momentum of the proton accelerated through 1.0 GV is $9.023\times {10}^{-19}\text{kg.}\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$.