QUANTUM PHYSICS FUNDAMENTALS - Chapter 4, Exercise 1 Solution
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De Broglie Wavelength from Momentum

PROBLEM
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An electron has a momentum of 5 x 10^-25 kg.m/s. Calculate its de
Broglie wavelength.

SOLUTION
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Using de Broglie's real formula:

    lambda = h / p

Substitute the given values:

    h = 6.626 x 10^-34 J.s
    p = 5 x 10^-25 kg.m/s

    lambda = (6.626 x 10^-34) / (5 x 10^-25)
    lambda = 1.325 x 10^-9 m (1.325 nm)

ANSWER: The electron's de Broglie wavelength is approximately
1.325 x 10^-9 m, or 1.325 nm.

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WHY THIS WORKS AS AN ANSWER

De Broglie's formula is a simple inverse proportionality between
wavelength and momentum - a slower-moving, lower-momentum electron
(as in this exercise, compared to the chapter's own faster 2 x 10^6
m/s worked example) has a correspondingly LARGER wavelength, since
dividing a fixed constant (h) by a smaller number gives a larger
result. This wavelength, at roughly 1.3 nm, is still comparable to
atomic and molecular scales, which is exactly why electrons at these
speeds continue to show real, observable wave behavior such as
diffraction.