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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