A free particle with initial kinetic energy E and de-broglie wavelength λ enters a region in which it has potential energy U. What is the particle's new de-Broglie wavelength?
A
λ(1−U/E)−1/2
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B
λ(1−U/E)
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C
λ(1−U/E)−1
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D
λ(1−U/E)1/2
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Solution
The correct option is Aλ(1−U/E)−1/2 Theinitialkineticenergyoffreeparticle:E=p22m→=√2mEDebrogliewavelength:λ=hp=h√2mEEnergyoftheparticlewhenitenterstheregion:Ef=E−USo,itswavelengthbecomes:λf=h√2mEf=h√2m(E−U)λ2f=h22mE(EE−V)=h22mE(11−U/E)λf=h√2mE(11−U/E)1/2=λ(1−U/E)−1/2