An electron and a proton are accelerated through a potential difference of 1.2 kV. What is the ratio of the de Broglie wavelength of the electron to the proton?

Question

BlackTom AI · Accepted Answer

Because both particles gain the same kinetic energy qV, the de Broglie wavelength scales as 1/√m; therefore λ_e/λ_p = √(m_p/m_e).

1) Relevant knowledge points and concepts
- When a charged particle is accelerated through a potential difference V, it gains kinetic energy K = qV (magnitude the same for electron and proton, q = elementary charge e).
- de Broglie wavelength: λ = h/p. For non-relativistic kinetic energy, p = √(2mK), so λ = h / √(2mK).
- With identical K for both particles, λ ∝ 1/√m.

2) Complete reasoning process step by step
- K_e = K_p = eV (here V = 1.2 kV). Thus K cancels when taking the ratio of wavelengths.
- λ_e / λ_p = [h / √(2 m_e K)] / [h / √(2 m_p K)] = √(m_p / m_e).

3) Calculations with intermediate steps
- Use masses: m_e = 9.10938356 × 10^−31 kg, m_p = 1.6726219 × 10^−27 kg.
- Mass ratio: m_p / m_e = (1.6726219e−27) / (9.10938356e−31) ≈ 1836.152673.
- Square root: √(1836.152673) ≈ 42.85.
- Rounding gives λ_e / λ_p ≈ 43.

4) Why the correct answer is correct
- Because both particles receive the same kinetic energy from the potential difference, their momenta differ only by √m. The much smaller electron mass gives it a much larger wavelength by the factor √(m_p/m_e) ≈ 43.

5) Why the other options are incorrect
- (a) 0.023 is approximately the reciprocal (1/43) and would correspond to λ_p/λ_e, not λ_e/λ_p.
- (c) 1.0 would imply equal masses or equal wavelengths, which is false.
- (d) 1800 is close to the mass ratio m_p/m_e (≈1836), not its square root; using that would ignore the square-root dependence in λ ∝ 1/√m.

6) Conclusion
Therefore the ratio λ_e/λ_p ≈ 43, matching the given correct answer (b. 43).

An electron and a proton are accelerated through a potential difference of 1.2 kV. What is the ratio of the de Broglie wavelength of the electron to the proton?单项选择题

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