The potential energy of a particle varies as V(x)=Eo;0≤x≤1 =0;x>1 For 0≤x≤1, de Broglie wavelength is γ1 and for x>1 the de Broglie wavelength is γ2.
Total energy of the particle is 2Eo. If γ1/γ2=√x. Find x
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Solution
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For 0≤x≤1, Kinetic energy E1=E0=p212m⇒p1=√2mE0 For x>1, E2=2E0=p222m⇒p2=√4mE0 now, γ1=hp1=h√2mE0 and γ2=hp2=h√4mE0 ∴γ1γ2=2√2=√2
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