Relationship between lambda and kinetic energy ?
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Hint: We will use the equation to find de-Broglie wavelength to find the kinetic energy of the electron with wavelength 1nm. De-Broglie wavelength of a particle is inversely proportional to the momentum of that particular body. We should know that kinetic energy and momentum of a particle is related as K. E=P22m.

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The kinetic energy of the electrons accelerated through a potential difference (voltage) V was E = ½mv2 = p2/(2m) = eV and the de Broglie formula then yields λ = h/(2meV)1/2, where e and m are the charge and the mass of the electron respectively.
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