alleryThe de Broglie wavelength of a particle of mass $m$ and momentum $p$ is
A$\lambda = h/p$
B$\lambda = hp/m$
C$\lambda = h \cdot p$
D$\lambda = mp/h$
Answer & Solution
Correct answer: A. $\lambda = h/p$
$\lambda = h/p$ is the de Broglie relation — every particle with momentum has an associated wavelength.
For an electron at $1\,eV$, this is about $1.2\,nm$
for a tennis ball, the wavelength is unmeasurably small.
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