i) Consider a particle of mass m and energy E in a 1-D square-well, with walls of finite height V(x| > a) = Vo > E, and V(|x) a, the Schrodinger equation can be written: 12 02x/ (2) 2m ax2 + ( Vo - E)y(x) = 0. [2 marks] ii) The general solution to this equation is v(|x| > a) = Ceat + De-at, where C, D, a are all constants. We will consider only the region where x > a, and can now simplify the general solution to v(x > a) = De-or. Show that: 2m(Vo - E) a = h2 Why can we ignore the Cet term in the trial wavefunction? [4 marks]