1.

Using Eq.(6.2e), find the probability `D` of a particle of mass `m` and energy `E` tunneling through the potential barrier shown in fig. where `U(x)=U_(0)(1-x^(2)//l^(2))`.

Answer» The potential is `U(x)=U_(0)(1-(x^(2))/(l^(2)))`. The turning points are
`(E )/(U_(0))=1-(x^(2))/(l^(2)) or x=+-lsqrt(1-(E )/(U_(0)))`
Then `D~~exp[-(4)/( ħ)int_(0)^(lsqrt(1-(E//U_(0))))sqrt(2m{U_(0))(1-(x^(2))/(l^(2)))-E}}dx]`
`=exp=exp[-(4)/(ħ)int_(0)^(lsqrt(1-(E//U_(0))))sqrt(2mU_(0))sqrt(1-(E)/(U_(0))-(x^(2))/(l^(2))dx)`
`=exp[-(4l)/(ħ)sqrt(2mV_(0)) int_(0)^(x_(0))sqrt(x_(0)^(2)-x^(2))dx], x_(0)=sqrt(1-E//V_(0))`
The intergal is `int_(0)^(x_(0))sqrt(x_(0)^(2)-x^(2))dx=x_(0)^(2)int_(0)^(x//2)cos^(2) thetad theta=(pi)/(4)x_(0)^(2)`
Thus `D~~exp[-(pil)/(ħ)sqrt(2mU_(0))(1-(E)/(U_(0)))]`
`=exp[-(pil)/(ħ)sqrt((2m)/(U_(0)))(U_(0)-E)]`


Discussion

No Comment Found

Related InterviewSolutions