Question

(b) Show that for certain incident energies there is 100 percent transmission. Suppose that we model an atom as a one-dimensional square well with a width of 1 A and that an electron with 0.7 eV of kinetic energy encounters the well. What must the depth of the well be for 100 percent transmission? This absence of scattering is observed when the target atoms are composed of noble gases such as krypton.

Answer 1

Step-by-step explanation:

100 persent transmission implies that the T=1

Therefore using the previous result we have

`1+\frac{\sin^2\sqrt{(2m)/(\hbar^2)(E+V_0)}a}{4(E)/(V_0)((E+V_0))/(V_0)}=1`

`\sin\sqrt{(2m)/(\hbar^2)(E+V_0)}a=0\Rightarrow \sqrt{(2m)/(\hbar^2)(E+V_0)}=0\Rightarrow E=-V_0`

The depth of the well for 100% transmission should be

`V_0=-0.7~{\rm{eV}}`