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Q5.12:
An electron and a proton are detected in a cosmic ray experiment, the first with kinetic energy 10 keV, and the second with 100 keV. Which is faster, the electron or the proton ? Obtain the ratio of their speeds. (electron mass = (9.11 \times 10^{-31}) kg, proton mass = (1.67 \times 10^{-27}) kg, 1 eV = (1.60 \times 10^{-19}) J).

Solution :

Step 1: Given Variables & Initial Setup

To determine the ratio of speeds, we identify the physical parameters provided:

  • Kinetic Energy of electron ($K_e$): $10 \text{ keV} = 10 \times 10^3 \text{ eV}$
  • Kinetic Energy of proton ($K_p$): $100 \text{ keV} = 100 \times 10^3 \text{ eV}$
  • Mass of electron ($m_e$): $9.11 \times 10^{-31} \text{ kg}$
  • Mass of proton ($m_p$): $1.67 \times 10^{-27} \text{ kg}$
  • Conversion factor: $1 \text{ eV} = 1.60 \times 10^{-19} \text{ J}$

[Per the Classical Kinetic Energy definition: $K = \frac{1}{2}mv^2$]

Step 2: Deriving the Expression for Speed

Rearranging the kinetic energy formula to isolate speed $v$:

$v = \sqrt{\frac{2K}{m}}$

To find the ratio of speeds ($v_e / v_p$), we establish the following ratio expression:

$\frac{v_e}{v_p} = \frac{\sqrt{\frac{2K_e}{m_e}}}{\sqrt{\frac{2K_p}{m_p}}} = \sqrt{\frac{K_e}{K_p} \times \frac{m_p}{m_e}}$

Step 3: Numerical Substitution

Substitute the given values into the ratio equation:

$\frac{v_e}{v_p} = \sqrt{\frac{10 \times 10^3}{100 \times 10^3} \times \frac{1.67 \times 10^{-27}}{9.11 \times 10^{-31}}}$

$\frac{v_e}{v_p} = \sqrt{\frac{1}{10} \times \frac{1.67}{9.11} \times 10^4}$

$\frac{v_e}{v_p} = \sqrt{0.1 \times 0.1833 \times 10^4} = \sqrt{183.3}$

$\frac{v_e}{v_p} \approx 13.54$

Step 4: Vector-Speed Comparison Visualization

0 v_e Electron Speed (v_e) is significantly higher Proton Speed (v_p)

Step 5: Analytical Conclusion

Since the ratio $\frac{v_e}{v_p} \approx 13.54$, which is greater than 1, we conclude that the electron is faster than the proton.

Final Solution: The electron is faster than the proton, and the ratio of their speeds is approximately 13.54:1.


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