The potential energy of a particle of mass m is given by U(x)=E0;0≤x≤10;x>1λ1 and λ2 are the de-Broglie wavelengths of the particle, when 0≤x≤1 and x>1 respectively. If the total energy of particle is 2E0, the ratioλ1λ2 will be
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a
2
b
1
c
2
d
12
answer is C.
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Detailed Solution
K.E. =2E0−E0=E0( for 0≤x≤1)⇒ λ1=h2mE0 K.E. =2E0( for x>1)⇒λ2=h4mE0⇒λ1λ2=2