What is the de Broglie wavelength associated with a neutron (mass 1.67 10^(-27) kg) moving at a speed of 2.0 10^3 m/s? (h = 6.63 10^(-34) Js)
Answer and explanation
Correct answer: A. 1.99 * 10^(-10) m (0.199 nm)
- Step 1: Recall the de Broglie wavelength formula: lambda = h / p = h / (m * v).
- Step 2: Calculate momentum: p = (1.67 10^(-27) kg) (2.0 10^3 m/s) = 3.34 10^(-24) kg*m/s.
- Step 3: Compute wavelength: lambda = (6.63 10^(-34) Js) / (3.34 10^(-24) kgm/s) approx 1.985 * 10^(-10) m.
Answer
1.99 * 10^(-10) m (0.199 nm).
Why not the others: '3.98 10^(-10) m' forgets the factor of 2 in velocity; '1.99 10^(-7) m' miscalculates powers of ten; '9.95 * 10^(-11) m' divides by 2.
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