What happens to the de Broglie wavelength of an electron if its momentum is doubled?

Question

BlackTom AI · Accepted Answer

The de Broglie relation λ = h/p relates a particle’s wavelength λ inversely to its momentum p, so increasing momentum decreases wavelength.

1) Relevant concepts:
- de Broglie wavelength: λ = h / p, where h is Planck’s constant and p is momentum.
- Inverse proportionality: if p changes by a factor, λ changes by the reciprocal factor.

2) Step-by-step reasoning:
- Let the initial momentum be p and the initial wavelength λ = h / p.
- If momentum is doubled, the new momentum p' = 2p.
- The new wavelength λ' = h / p' = h / (2p).

3) Intermediate calculation steps:
λ' = h / (2p) = (1/2) × (h / p) = (1/2) × λ.
Hence λ' / λ = 1/2, so the wavelength is reduced to one-half its original value.

(As a quick numeric check: choose p = 1.0 × 10^-24 kg·m/s and h ≈ 6.626×10^-34 J·s.
λ = 6.626×10^-34 / 1.0×10^-24 = 6.626×10^-10 m.
With p' = 2.0×10^-24, λ' = 6.626×10^-34 / 2.0×10^-24 = 3.313×10^-10 m = λ/2.)

4) Why this is correct:
Because λ is inversely proportional to p, doubling p multiplies the denominator by 2, halving the entire fraction. Thus the wavelength decreases by a factor of 2.

5) Why the other options are incorrect:
- a (decreases by factor of 4): would require momentum to increase by factor 4 (λ ∝ 1/p), not 2.
- b (increases by factor of 2) and c (increases by factor of 3) and e (increases by factor of 4): increasing momentum cannot increase wavelength because of the inverse relationship; these are the opposite of the correct behavior.

6) Conclusion:
Therefore the de Broglie wavelength decreases by a factor of 2 when the electron’s momentum is doubled. This confirms answer d.

What happens to the de Broglie wavelength of an electron if its momentum is doubled?单项选择题

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