What is the de Broglie wavelength of an electron accelerated through a potential difference of 100 V? (Take h / sqrt(2me) = 1.227 nm V^(1/2))

Answer and explanation

Correct answer: A. 0.123 nm (1.23 Angstroms)

  1. Step 1: De Broglie wavelength for electron: lambda = 1.227 / sqrt(V) nm. Step 2. For V = 100 V: sqrt(100) = 10. Step 3. lambda = 1.227 / 10 = 0.1227 nm = 0.123 nm. Answer: 0.123 nm (1.23 Angstroms). Why not the others: 1.227 nm is for 1 V; 0.012 nm is for 10,000 V.

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