Asked by MIke
In a Rutherford scattering experiment a target nucleus has a diameter of 1.4 10-14 m. The incoming has a mass of 6.64 10-27 kg. What is the kinetic energy of an particle that has a de Broglie wavelength equal to the diameter of the target nucleus? Ignore relativistic effects
The momentum is
p = h/lambda
Kinetic energy is:
p^2/(2m) = (h/lambda)^2/(2m) =
8.43*10^(-14)J = 0.526 MeV.
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The momentum is
p = h/lambda
Kinetic energy is:
p^2/(2m) = (h/lambda)^2/(2m) =
8.43*10^(-14)J = 0.526 MeV.
nvjhjvjhcvjv
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