Question

Making use of the solution of the foregoing problem, determine the probability of the particle with energy `E=U_(0)//2` to be located in the region `xgtl`, if `l^(2)U_(0)=((3)/(4)pi)^(2)( ħ^(2))/(m)`.

Answer 1

`U_(0)l^(2)=((3)/(4)pi)U_(0)l^(2)=((3)/(4)pi)^(2)( ħ^(2))/(m)` and `El^(2)=((3)/(4)pi)^(2)( ħ^(2))/(2m)`
or `kl=(3)/(4)pi`
It is easy to check that the condition of the bound state is satisfied. Also
`alphal=sqrt((2m)/ħ^(2)(U_(0-E))l^(2))= sqrt((mU_(0))/ħ^(2)l^(2))=(3)/(4)pi`
Then from the previous problem
`D=Ae^(alphal)sin kl=A(e^(3//4))/(sqrt(2))`
By normalization
`I=A^(2)[ int_(0)^(l)sin^(2)kxdx+int_(l)^(oo)(3^(3pi//2))/(2)e^(-(3pi//2)x//l)dx]`
`A^(2)[(1)/(2)int_(0)^(1)(1-cos 2kx)dx+1int_(0)^(oo)(1)/(2)e^(-3x)/(2)ydy]`
`=A^(2)[(1)/(2)[-(sin 2kl)/(2k)]+(1)/(2).((l)/(3pi))/(2)]=A^(2)l[(1)/(2)[1+((1)/(3pi))/(2)]+(1)/(2)((1)/(3pi))/(2)]`
`A^(2)l[(1)/(2)+(2)/(3pi)]=A^(2)(l)/(2)(1+(4)/(3pi))` or `A =sqrt((2)/(l))(1+(4)/(3pi))^(-1//2)`
The probability of the particle to be located in the region `x gt l` is
`P=int_(l)^(oo)Psi^(2)dx=(2)/(l)(1+(4)/(3pi))^(-1)int_(l)^(oo)(e^(3)pi//2)/(2) e^(-(3 x)/(2)(x)/(l))dx`
`=(1+(4)/(3pi))^(-1)int_(l)^(oo)e^(3pi//2)e^(-(3pi//2)y)dy=(2)/(3pi)xx(3pi)/(3pixx4)=14.9%`.