Question

An electron and proton possess same K.E., how will their momentum be related?

Answer 1

Step 1: Given data and assumptions

Let the kinetic energy of a proton be, $K{E}_p$

Let, the kinetic energy of the electron will be, $K{E}_e$

Step 2: Obtaining a relationship between the momentum of a proton and an electron

The relation between kinetic energy and momentum can be expressed as,

$KE=\frac{p^2}{2m}$

where $KE$ is kinetic energy, $p$ is momentum and $m$ is mass.

Now, according to the question, an electron and a proton possess the same kinetic energy. Therefore,

$$\begin{gathered} K{E}_e=K{E}_p \\ \frac{{p_e}^2}{2m_e}=\frac{{p_p}^2}{2m_p} \\ \frac{{p_e}^2}{m_e}=\frac{{p_p}^2}{m_p} \end{gathered}$$

The above equation $p_e$ represents the momentum of the electron, $m_e$ represents the mass of the electron, $p_p$ represents the momentum of the proton and $m_p$ represents the mass of the proton.

Now, the mass of an electron is, $m_e=9.1\times {10}^{31}kg$ and the mass of a proton is, $1.67\times {10}^{-27}kg$

Substituting these values in the above equation, we get,

$$\begin{gathered} \frac{{p_e}^2}{9.1\times {10}^{31}kg}=\frac{{p_p}^2}{1.67\times {10}^{-27}kg} \\ {p_p}^2=\frac{1.67\times {10}^{-27}}{9.1\times {10}^{31}}\times {p_e}^2 \\ {p_p}^2=\frac{1.67\times {10}^{-27}}{9.1\times {10}^{31}}\times {p_e}^2 \\ {p_p}^2=1.8\times {p_e}^2 \\ {p}_p=1.34p_e \end{gathered}$$

Therefore, the momentum of a proton is approximately equal to $1.34$ times the momentum of an electron.