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Question

# The potential energy of particle of mass m varies as U(x)={E0: 0≤x≤10: x>1 The de Broglie wavelength of the particle in the range 0≤x≤1isλ1 and that in the range x>1isλ2. If the total energy of the particle is 2E0, find λ1,/λ2.

A
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C
1
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D
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Solution

## The correct option is A √2de Broglie wavelength of a particle of ///λ=h√2mKwhere K is the kinetic energy of the particle. For 0≤x≤1,U(x)=E0As, total energy = Kinetic energy + potential energy∴2E0=K1+E0or K1=E0∴λ1=h√2mK1=h√2mE0For x>1,U(x)=0∴K2=2E0∴λ2=h√2mK2=h√2m(2E0)Divide (i) by (ii), we get λ1λ2=h√2mE0×√2m(2E0)h=√2

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