Determine the de Broglie wavelength of a neutron (m = 1.67 x 10-27kg) which has a speed of 5.0 m/s.

Question

BlackTom AI · Accepted Answer

λ = h / (mv), so use Planck’s constant h = 6.626 × 10^-34 J·s.

1) Relevant concepts:
- de Broglie wavelength λ = h/p, where p is momentum.
- For non-relativistic particles p = mv (v << c here, so classical momentum is valid).
- Unit conversion: 1 m = 10^9 nm.

2) Step-by-step calculation:
- Given m = 1.67 × 10^-27 kg and v = 5.0 m/s.
- Compute momentum p = m v = (1.67 × 10^-27 kg)(5.0 m/s) = 8.35 × 10^-27 kg·m/s.
- Compute wavelength λ = h / p = (6.626 × 10^-34 J·s) / (8.35 × 10^-27 kg·m/s).

3) Intermediate arithmetic:
- 6.626 × 10^-34 ÷ 8.35 × 10^-27 = (6.626/8.35) × 10^(-34+27)
- 6.626/8.35 ≈ 0.793
- Exponent: 10^(-7)
- Therefore λ ≈ 0.793 × 10^-7 m = 7.93 × 10^-8 m.

4) Convert to nanometers:
- 7.93 × 10^-8 m × (10^9 nm / 1 m) = 7.93 × 10^1 nm = 79.3 nm.
- Rounding to two significant figures (consistent with given data) gives ≈ 79 nm.

5) Why the correct answer is correct:
- The computed de Broglie wavelength (≈79.3 nm) directly matches choice (a) 79 nm using the standard formula and given values. The particle speed is non-relativistic, so no relativistic correction is needed.

6) Why other options are incorrect (brief):
- (b) 162 nm and larger values (c–e) are significantly higher than the computed 79.3 nm and would require a much smaller momentum (or much smaller mass/speed) than given. They do not match the arithmetic above.

Conclusion: The de Broglie wavelength is about 79 nm, confirming answer (a) 79 nm.

Determine the de Broglie wavelength of a neutron (m = 1.67 x 10-27kg) which has a speed of 5.0 m/s.单项选择题

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