What is the de Broglie wavelength associated with an electron, accelerated through a potential difference of 200 volts?
Correct answer: D. 0.086 nm
- A. 1 nm
- B. 0.5 nm
- C. 0.0056 nm
- D. 0.086 nm
Explanation
λ = h/pWe can find the momentum of the electron from the potential difference:qV = 1/2 mv^2 = p^2/(2m)Solving for p, we get:p = sqrt(2mqV)Substituting this value of p in the de Broglie wavelength formula, we get:λ = h/sqrt(2mqv)where m is the mass of the electron.Substituting the given values, we get:λ = 6.626 x 10^-34 J.s / sqrt(2 x 9.109 x 10^-31 kg x 1.602 x 10^-19 C x 200 V)λ ≈ 0.086 nm.
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Classical physics cannot explain blackbody radiation and related observations, leading to Planck’s quantum theory, in which energy is emitted or absorbed in discrete packets. Photons provide the particle model of light, with energy and momentum linked to frequency and wavelength, including the photoelectric effect and its threshold frequency.
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