Solution:
The de Broglie wavelength of an electron is \(\lambda = \frac{1.227}{\sqrt{V}} \text{nm}\). Substituting \(V = 81 \text{V}\), we get \(\lambda = \frac{1.227}{9} \approx 0.136 \text{nm}\).
The de Broglie wavelength associated with an electron, accelerated by a potential difference of 81 V is given by:
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