alleryFor accelerated electron with KE = 100 eV, de Broglie wavelength:
A1.23 Å ≈ 0.123 nm (= 12.27/sqrt(100))
B100 Å
C1 nm
D0.01 nm
Answer & Solution
Correct answer: A. 1.23 Å ≈ 0.123 nm (= 12.27/sqrt(100))
λ ≈ 12.27/sqrt(V) Å where V in volts. V = 100: λ ≈ 1.23 Å.
Comparable to atomic spacing; gives observable electron diffraction.
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