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Particle of mass m and energy E moves from the left to the potential barrier shown in figure. Find: (a) the reflection coefficient R of the barrier for E gt U_(0) (b) the effective penetration depth of the particles into the region x gt 0 for E lt U_(0)i.e., the distacne from the barrier boundary to the point at which the probability of finding a particle decreases e-fold |
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Answer» Solution :(a) We start from the Schrodinger equation `(d^(2)Psi)/(dx^(2))+(2m)/( ħ^(2))(E-U(x))Psi=0` which we write as `Psi_(1)''+k^(2)Psi_(1)=0, x LT 0` `k^(2)=(2mE)/( ħ^(2))` and `Psi_(II)+alpha^(2)Psi_(II)=0x GT 0` `alpha^(2)=(2m)/( ħ^(2))(E-U_(0)) gt 0` It is convinient to look for solutions in the form `Psi_(I)=e^(ikx)+Re^(-ikx)x lt 0` `Psi_(II)=Ae^(i alphax)+Be^(-alpha x)x gt 0` In region (I)`(x lt 0)`, the amplitude of `e^(ikx)` is written as unity by conservation. In II we EXPECT only a transmitted wave to the right, `B=0` then. So `Psi_(II)=Ae^(ikx)x gt 0` The boundary conditions follow from the continuity of `Psi(dPsi)/(dx) at x=0` `1+R=A` `iK(1-R)= i alpha A` Then `(1-R)/(1+R)=(alpha)/(k)` or `R=(k-alpha)/(k+alpha)` The reflection coefficient is the absolute square of `R`: `r=|R|^(2)=|(k-alpha),(k+alpha)|^(2)` (b) In this case `E lt U_(0), alpha^(2)= -beta^(2) lt 0`. Then `Psi_(1)` is uncharged in from but `Psi_(II)=Ae^(-betax)+Be^(+betax)` we must have `B=0` since otherwise `Psi(x)` will become unbounded as `xrarroo`. INSIDE the barrier, the PARTICLE then has probability density equal to `|Psi_(II)|^(2)=|A|^(2)e^(-2betax)` This decreases to `(1)/(e )` of its value in `x_(eff)=(1)/(2beta)=( ħ)/(2sqrt(2m(U_(0)-E)))` |
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