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Employing Eq.(6.2e), find the probability D of an electron with energy E tunneling through a potential barrier of width l and height U_(0) provided the barrier is shaped as shown: (a) In fig 6.4 (b) in figure 6.5. |
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Answer» Solution :The formula is `D~~exp[-(2)/( ħ)int_(x_(1))^(x_(2))SQRT(2m(V(x)-E)dx)]` Here `V(x_(2))=V(x_(1))=E` and `V(x)gtE` in the region `x_(2)gt xgt x_(1)`. (a) For the problem, the intergral is trivial `D~~exp[-(2L)/( ħ)sqrt(2m(U_(0)-E))]` (b) We can without loss of GENERALITY take `x=0` at the point the potential begins to CLIMB. Then `U(x)= ,{{:(0x,LT,0,,),(U_(0),(x)/(l),0 x lt l,,),(0,xgtl,,,):}` Then `D~~exp[-(2)/ (ħ)int_(l(E)/(U_(0)))^(l)sqrt(2m(U_(0)(x)/(l)-E)dx)]` `=exp[-(2)/ (ħ)sqrt((2mU_(0))/(l))^(l)int_(x_(0))^(l)sqrt(x-x_(0))dx]x_(0)=l(E )/(U_(0))` `=epx[-(2)/( ħ)sqrt((2mU_(0))/(l))(2)/(3)(x-x_(0))^(3//2):|_(x_(0))^(l)]` `=exp[-(4)/(3 ħ)sqrt((2mU_(0))/(l))(l-l(E )/(U_(0)))^(3//2)]` `= exp[-(4l)/( 3ħU_(0))(U_(0)-E)^(3//2)sqrt(2m`)] |
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