Word Count : 204

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The de Broglie wavelength of an electron is given by [ \lambda = \frac{h}{p} = \frac{h}{\sqrt{2 m_e e V}} ] where (h = 6.626 \times 10^{-34} \text{ Js}), (m_e = 9.11 \times 10^{-31} \text{ kg}), (e = 1.602 \times 10^{-19} __________ ________ ___ _________ ________ ________ __________ _____ ____ ___.
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