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An electron of momentum $p = 6.626\times 10^{-24}$ kg m/s has de Broglie wavelength:
A$6.626\times 10^{-24}$ m, identical to the momentum value
B$10^{-24}$ m, ignoring Planck's constant entirely
C$10^{10}$ m, since wavelength scales hugely above the value
D$10^{-10}$ m, since $\lambda = h/p = 6.626\times 10^{-34}/p$
Answer & Solution
Correct answer: D. $10^{-10}$ m, since $\lambda = h/p = 6.626\times 10^{-34}/p$
$\lambda = h/p = 6.626\times 10^{-34}/6.626\times 10^{-24} = 10^{-10}$ m.
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