- #1

Kynsuo

- 14

- 1

- Homework Statement
- Show that psi is a solution to the radial component of the TISE for the electron in the Hydrogen Atom, find the value of ##a_0## in terms of fundamental constants and find the energy for this state.

- Relevant Equations
- $$- \frac{\hbar^2}{2mr^2} \frac d {dr} \left(r^2 \frac {d \psi}{dr} \right) + V(r) \psi = E \psi$$

$$V(r) = -\dfrac 1 {4 \pi \varepsilon_0} \dfrac{e^2}{r}$$

$$\psi(r) = \dfrac 1 {\sqrt{\pi a_0^3}} \exp\left(-\dfrac r {a_0}\right)$$

I am unable to complete the first part of the question. After I plug in the function for psi into the differential equation I am stuck:

$$\frac {d \psi (r)}{dr} = -\frac 1 a_0 \psi (r), \frac d{dr} \biggl(r^2 \frac {d\psi (r)}{dr} \biggr) = -\frac 1 {a_0}\frac d {dr} \bigl[r^2 \psi(r) \bigr] = - \frac 1 {a_0}\biggl[ 2r\psi(r) - \frac{r^2}{a_0} \psi(r) \biggr]$$

$$\frac{\hbar }{2mr^2} \frac 1 {a_0} \biggl[ 2r \psi(r) - \frac {r^2}{a_0} \psi(r) \biggr] + V(r) \psi(r) = E \psi(r) $$

$$\Leftrightarrow \psi (r) \biggl[ \frac {\hbar}{2mr^2} \frac {2ra_0-r^2}{a_0^2} - \frac{e^2}{4 \pi \varepsilon_0 r} \biggr] = E \psi(r)$$

I don't the how the expression I get in the brackets in the RHS at the end is equal to the total energy.

$$\frac {d \psi (r)}{dr} = -\frac 1 a_0 \psi (r), \frac d{dr} \biggl(r^2 \frac {d\psi (r)}{dr} \biggr) = -\frac 1 {a_0}\frac d {dr} \bigl[r^2 \psi(r) \bigr] = - \frac 1 {a_0}\biggl[ 2r\psi(r) - \frac{r^2}{a_0} \psi(r) \biggr]$$

$$\frac{\hbar }{2mr^2} \frac 1 {a_0} \biggl[ 2r \psi(r) - \frac {r^2}{a_0} \psi(r) \biggr] + V(r) \psi(r) = E \psi(r) $$

$$\Leftrightarrow \psi (r) \biggl[ \frac {\hbar}{2mr^2} \frac {2ra_0-r^2}{a_0^2} - \frac{e^2}{4 \pi \varepsilon_0 r} \biggr] = E \psi(r)$$

I don't the how the expression I get in the brackets in the RHS at the end is equal to the total energy.