What is the kinetic energy of a proton moving with speed 3.4 × 103 m s−1, in electronvolts?

Question

BlackTom AI · Accepted Answer

Kinetic energy is KE = (1/2) m v^2, so use the proton mass and given speed.

1) Relevant knowledge:
- Classical kinetic energy formula KE = 1/2 m v^2.
- Proton mass m = 1.6726219 × 10^−27 kg.
- 1 electronvolt (eV) = 1.602176634 × 10^−19 J. Converting J → eV: divide by 1.602×10^−19.

2) Step-by-step calculation with intermediate steps:
- v = 3.4 × 10^3 m/s, so v^2 = (3.4 × 10^3)^2 = 11.56 × 10^6 = 1.156 × 10^7 m^2/s^2.
- m v^2 = (1.6726219 × 10^−27 kg) × (1.156 × 10^7 m^2/s^2)
         = 1.6726219 × 1.156 × 10^(−27+7)
         ≈ 1.93355 × 10^−20 J.
- KE = 1/2 m v^2 = 0.5 × 1.93355 × 10^−20 J ≈ 9.66775 × 10^−21 J.

3) Convert to electronvolts:
- KE (eV) = (9.66775 × 10^−21 J) / (1.602176634 × 10^−19 J/eV)
          ≈ (9.66775 / 1.602176634) × 10^(−21+19)
          ≈ 6.034 × 10^−2 eV
          = 0.06034 eV = 60.34 meV.

4) Why this answer is correct:
- The numeric result 0.0603 eV equals about 60 meV, which matches the given correct choice when rounded to two significant figures. The calculation used standard constants and straightforward unit conversion.

5) Why the other choices are incorrect:
- "9.7 × 10^−21 eV" is roughly our KE in joules (9.67 × 10^−21 J), but with the wrong unit; as an energy in eV it would be many orders of magnitude too small.
- "33 μeV" = 33 × 10^−6 eV = 3.3 × 10^−5 eV, far smaller than 0.0603 eV.
- "35 μeV" likewise (3.5 × 10^−5 eV) is much smaller than the computed value.

Conclusion: The correct kinetic energy is approximately 60 meV, which confirms the provided answer.

What is the kinetic energy of a proton moving with speed 3.4 × 103 m s−1, in electronvolts?单项选择题

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