The average distance of the electron from the proton in the hydrogen atom is 0.51 x 10-10 m. What is the electric field from the proton’s charge at the location of the electron? (ke = 8.99 x 109 N×m2/C2, elementary charge e = 1.6 x 10-19 C)

Question

The average distance of the electron from the proton in the hydrogen atom is 0.51 x 10-10 m. What is the electric field from the proton’s charge at the location of the electron? (ke = 8.99 x 109 N×m2/C2,  elementary charge e = 1.6 x 10-19 C)

BlackTom AI · Accepted Answer

Electric field of a point charge: E = ke q / r^2, so use ke = 8.99×10^9 N·m^2/C^2, q = e = 1.6×10^-19 C, r = 0.51×10^-10 m.

1) Relevant knowledge points and concepts
- Coulomb’s law / electric field of a point charge: magnitude E = ke·q / r^2.
- Units: ke (N·m^2/C^2) × q (C) / r^2 (m^2) → N/C (correct units for E).
- The proton is a point positive charge; the field magnitude at the electron’s average radius is what we compute.

2) Complete reasoning process (step by step)
- Convert the given radius to standard scientific notation: r = 0.51×10^-10 m = 5.1×10^-11 m.
- Compute r^2, compute numerator ke·q, then divide to get E.

3) Calculations with intermediate steps
- r = 5.1×10^-11 m
- r^2 = (5.1×10^-11 m)^2 = 26.01×10^-22 m^2 = 2.601×10^-21 m^2
- ke·q = (8.99×10^9 N·m^2/C^2) × (1.6×10^-19 C) = 14.384×10^-10 N·m^2/C = 1.4384×10^-9 N·m^2/C
- E = (1.4384×10^-9) / (2.601×10^-21) N/C = (1.4384 / 2.601) ×10^(12) N/C ≈ 0.5529×10^12 N/C
- E ≈ 5.53×10^11 N/C, which to two significant figures is 5.5×10^11 N/C.

4) Why the correct answer is correct
- The numerical result follows directly from the point-charge formula with correct substitution and unit handling. The magnitude is large because r is extremely small (~10^-11 m), making r^2 very small and thus E very large.

5) Why the other multiple-choice options are incorrect
- 3.2×10^2, 8.8×10^-8, 1.4×10^6, and 4.4×10^9 N/C are many orders of magnitude smaller than the correct value; they do not match the r^-2 dependence for the given tiny radius and therefore are inconsistent with direct calculation.

6) Conclusion
- The calculated electric field magnitude at the electron’s average distance is ≈5.5×10^11 N/C, matching the provided correct answer.

The average distance of the electron from the proton in the hydrogen atom is 0.51 x 10-10 m. What is the electric field from the proton’s charge at the location of the electron? (ke = 8.99 x 109 N×m2/C2, elementary charge e = 1.6 x 10-19 C)单项选择题

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