A nonrelativistic electron is accelerated from rest through a potential difference.After acceleration the electron has a de Broglie wavelength of 880 nm.What is the potential difference through which this electron was accelerated? (h = 6.626 × 10^-34 J • s,e = 1.60 × 10^-19 C,m_el = 9.11 × 10^-31 kg) A) 1.9 µV B) 1.7 µV C) 2.2 µV D) 2.5 µV

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The de Broglie wavelength is given by:$$\lambda = \frac{h}{p}$$where h is the Planck constant, p is the momentum of the electron, and m is the mass of the electron.The momentum of the electron is given by:$$p = mv$$where v is the velocity of the electron.The kinetic energy of the electron is given by:$$K = \frac{1}{2}mv^2$$where K is the kinetic energy of the electron.The potential difference through which the electron was accelerated is given by:$$V = K/e$$where V is the potential difference, e is the charge of the electron, and K is the kinetic energy of the electron.Substituting the expressions for p, v, and K into the expression for V, we get:$$V = \frac{h^2}{2me\lambda^2}$$Substituting the given values into the expression for V, we get:$$V = \frac{(6.626 	imes 10^{-34} 	ext{ J} \cdot 	ext{s})^2}{2(9.11 	imes 10^{-31} 	ext{ kg})(1.60 	imes 10^{-19} 	ext{ C})(880 	imes 10^{-9} 	ext{ m})^2}$$$$V = 1.9 \mu	ext{V}$$Therefore, the potential difference through which the electron was accelerated is 1.9 µV.

A Nonrelativistic Electron Is Accelerated from Rest Through a Potential