Matter waves: 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 though 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

Question

Quizplus · Accepted Answer

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 velocity of the electron is given by:$$v = \sqrt{\frac{2eV}{m}}$$where e is the charge of the electron, and V is the potential difference through which the electron was accelerated.Substituting these equations into the de Broglie wavelength equation, we get:$$\lambda = \frac{h}{\sqrt{2eV}}$$Solving for V, we get:$$V = \frac{h^2}{2e\lambda^2}$$Substituting the given values, we get:$$V = \frac{(6.626 	imes 10^{-34} 	ext{ J} \cdot 	ext{s})^2}{2(-1.60 	imes 10^{-19} 	ext{ C})(880 	imes 10^{-9} 	ext{ m})^2} = 1.9 \mu	ext{V}$$Therefore, the potential difference through which the electron was accelerated is 1.9 µV.

Matter Waves: a Nonrelativistic Electron Is Accelerated from Rest Through