The potential energy of a particle of mass m is given by V(x)={E000≤x≤1x>1}
λ1 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, find (λ1/λ2).
Total energy , E=T+V where T=Kinetic energy and V=potential energy.
When 0≤x≤1, 2E0=T+E0⇒T=E0
de-Broglie wavelength, λ1=h√2mT=h√2mE0
When x>1,2E0=T+0 or T=2E0
now, λ2=h√2mT=h√2m2E0
Thus, λ1λ2=√2