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

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

A Nonrelativistic Electron Is Accelerated from Rest Through a Potential