what is the relation between kimetic energy and de-brogiles wavelength
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If you double the kinetic energy of a particle, how does the deBroglie wavelength change? Solution: λ = h/p, E = p2/(2m), p is proportional to √E, l is proportional to 1/√E. λ2/λ1 = √(E1/E2) = 1/√2.
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If you double the kinetic energy of a particle, how does the deBroglie wavelength change? Solution: λ = h/p, E = p2/(2m), p is proportional to √E, l is proportional to 1/√E. λ2/λ1 = √(E1/E2) = 1/√2.
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