1.

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.

Answer» 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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