A proton in a linear accelerator has a de Broglie wavelength of 141 pm .

Question

What is the speed of the proton? (The mass of the proton is 1.673×10−271.673×10−27 kgkg and 1 JJ = 1 (kg⋅m2)/s2(kg⋅m2)/s2.)

Answer 1

v = 2.81 × 10³ m/s (about 2810 m/s)

The relation

λ = h / (mv)  →  v = h / (mλ)

Step 1 — convert the wavelength to metres

141 pm = 141 × 10⁻¹² m = 1.41 × 10⁻¹⁰ m

Step 2 — substitute

v = (6.626 × 10⁻³⁴ kg·m²/s) / [(1.673 × 10⁻²⁷ kg)(1.41 × 10⁻¹⁰ m)]

Denominator: (1.673 × 10⁻²⁷)(1.41 × 10⁻¹⁰) = 2.359 × 10⁻³⁷ kg·m

v = (6.626 × 10⁻³⁴) / (2.359 × 10⁻³⁷) = 2.81 × 10³ m/s

Why the problem hands you "1 J = 1 kg·m²/s²"

That hint exists so you can see the units resolve. Planck's constant is in J·s, and J·s = (kg·m²/s²)·s = kg·m²/s. Divide by kg·m and you are left with m/s — a speed. If your units do not collapse to m/s, you have made a substitution error somewhere, and checking this takes ten seconds.

Worth noticing about the answer

2810 m/s sounds fast, but it is only about 0.001% of the speed of light. The phrase "linear accelerator" primes you to expect something relativistic, and it is reasonable to wonder whether the simple formula still applies here. It does, comfortably — relativistic corrections do not matter until you are a few percent of c, so the non-relativistic de Broglie relation is entirely appropriate.

The underlying reason the speed is modest: a proton is about 1800 times heavier than an electron. For the same wavelength, momentum is fixed at h/λ, so a heavier particle needs proportionally less speed. An electron with this same 141 pm wavelength would be moving around 5.2 × 10⁶ m/s.

Two common slips

1. Leaving λ in picometres. That throws the answer off by 10¹².
2. Inverting the formula. Solving λ = h/mv for v means dividing h by mλ, not multiplying. If your answer comes out absurdly small, check this first.