An electron has de Broglie wavelength 2.76Ă—10^(-10) m. Part A: Determine the magnitude of its momentum. Part B: Determine its kinetic energy in joules. Part C: Determine its kinetic energy in electron volts.
Added by Madeline V.
Step 1
626 x 10^-34 Js) and p is the momentum of the particle. We can rearrange this equation to solve for p: p = h/λ. Substituting the given values, we get p = (6.626 x 10^-34 Js) / (2.76 x 10^-10 mm) = 2.4 x 10^-24 kg m/s. Show more…
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