Answer:
Explanation:
100 persent transmission implies that the T=1
Therefore using the previous result we have
[tex]1+\frac{\sin^2\sqrt{\frac{2m}{\hbar^2}(E+V_0)}a}{4\frac{E}{V_0}\frac{(E+V_0)}{V_0}}=1 [/tex]
[tex]\sin\sqrt{\frac{2m}{\hbar^2}(E+V_0)}a=0\Rightarrow \sqrt{\frac{2m}{\hbar^2}(E+V_0)}=0\Rightarrow E=-V_0 [/tex]
The depth of the well for 100% transmission should be
[tex]V_0=-0.7~{\rm{eV}}[/tex]