QUANTUM PHYSICS FUNDAMENTALS - Chapter 4, Exercise 2 Solution
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De Broglie Wavelength of a Proton

PROBLEM
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A proton (mass 1.673 x 10^-27 kg) moves at 3 x 10^5 m/s. Calculate
its momentum, then its de Broglie wavelength.

SOLUTION
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Step 1: Calculate momentum.

    p = m v
    p = (1.673 x 10^-27) x (3 x 10^5)
    p = 5.019 x 10^-22 kg.m/s

Step 2: Calculate the de Broglie wavelength.

    lambda = h / p
    lambda = (6.626 x 10^-34) / (5.019 x 10^-22)
    lambda = 1.32 x 10^-12 m (approximately)

ANSWER: The proton's momentum is approximately 5.02 x 10^-22 kg.m/s,
and its de Broglie wavelength is approximately 1.32 x 10^-12 m.

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WHY THIS WORKS AS AN ANSWER

Even though this proton moves slower than the chapter's own electron
example (3 x 10^5 m/s vs. 2 x 10^6 m/s), its own real wavelength
comes out roughly 1,000 times SMALLER than the electron's own
1.325 x 10^-9 m result from Exercise 1 - not larger. This is because
the proton's mass (1.673 x 10^-27 kg) is roughly 1,836 times greater
than the electron's mass, and momentum depends on mass as well as
velocity. The proton's own much larger mass dominates the
calculation, giving it a much larger momentum and therefore a much
smaller de Broglie wavelength - a real, direct illustration of why
heavier particles show far weaker, less easily observed wave
behavior than lighter ones like electrons, even at comparable or
even greater real speeds.