# Kinetic Energy and the Schrodinger Equation

Let $$\psi_1(r), \psi_2(r)$$ be two wave-functions for an electron, which are plotted in the graph below. It may be observed of $$\psi_1(r)$$, that it possesses greater curvature than $$\psi_2(r)$$. If, by the Schrodinger Equation, $$-\frac{h^2}{2m}\frac{d^2\psi}{dx^2} = K\psi$$, may it be concluded that the kinetic energy of the first particle (i.e., that corresponding to $$\psi_1(r)$$) exceeds the second ($$\psi_2(r)$$)?

I am reserved on this answer, for the kinetic energy does not strictly depend upon the curvature of the graph, but also on the height of the graph (the value of $$\psi(x)$$), in such a way that if the height should be larger then the kinetic energy is smaller.

Edit: The potential energy graph applies for both particles.

• $\psi$'s are positive real functions for small $r$ according to the graphics. The curvatures are also positive. Is the energy $K$ negative to fulfill the SE? – Claudio Saspinski Aug 28 '20 at 22:25