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MCQ
The de Broglie wavelength of an electron accelerated through 100 V is approximately:
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A 0.123 nm
B 1.23 nm
C 0.012 nm
D 12.3 nm
Explanation

λ = h/√(2meV) = 1.227/√V nm. For V = 100, λ = 1.227/10 = 0.1227 nm ≈ 0.123 nm

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