Q.An electron and a proton are detected in a cosmic ray experiment, the first with kinetic energy $10\ \text{keV}$, and the second with $100\ \text{keV}$. Which is faster, the electron or the proton? Obtain the ratio of their speeds. (electron mass $= 9.11 \times 10^{-31}\ \text{kg}$, proton mass $= 1.67 \times 10^{-27}\ \text{kg}$, $1\ \text{eV} = 1.60 \times 10^{-19}\ \text{J}$).

Concept: Kinetic Energy Speed Relation — For a particle of mass $m$ and kinetic energy $K$, the speed is $v = \sqrt{2K/m}$.

Step 1: Convert energies to joules.

$K_e = 10\ \text{keV} = 10 \times 10^3 \times 1.60 \times 10^{-19} = 1.60 \times 10^{-15}\ \text{J}$

$K_p = 100\ \text{keV} = 100 \times 10^3 \times 1.60 \times 10^{-19} = 1.60 \times 10^{-14}\ \text{J}$

Step 2: Write speed ratio.

$\frac{v_e}{v_p} = \sqrt{\frac{K_e}{K_p} \cdot \frac{m_p}{m_e}} = \sqrt{\frac{1.60 \times 10^{-15}}{1.60 \times 10^{-14}} \cdot \frac{1.67 \times 10^{-27}}{9.11 \times 10^{-31}}}$

Step 3: Simplify. …

The core idea here is simple: kinetic energy depends on both mass and speed. When two particles have different masses but comparable kinetic energies, the lighter one must be moving much faster. This is a direct consequence of $K = \frac12 m v^2$.

Let’s work through it step by step.

1. Write the kinetic energy relation for each particle.

   For the electron: $K_e = \frac12 m_e v_e^2$

   For the proton: $K_p = \frac12 m_p v_p^2$

2. We want the ratio of speeds $v_e / v_p$.

   Divide the two equations:

3. Rearrange to isolate the speed ratio.

Taking square roots:

4. Plug in the numbers. $K_e = 10\ \text{keV}$, $K_p = 100\ \text{keV}$, so $K_e / K_p = 0.1$ $m_e = 9.11 \times 10^{-31}\ \text{kg}$, $m_p = 1.67 \times 10^{-27}\ \text{kg}$

Therefore:

$\frac{v_e}{v_p} = \sqrt{0.1 \times 1833} = \sqrt{183.3} \approx 13.54$ …

Concept: Kinetic Energy – Speed Relation

Step 1: Write kinetic energy for each particle

Step 2: Form the ratio of speeds

Step 3: Substitute values …