- #1

TFM

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## Homework Statement

The groundstate energy of the hydrogen atom can be calculated using the variational

principle. The normalised groundstate wavefunction is:

[tex] \psi_{100} = R(r)_{10} \cdot Y_{00} [/tex]

with [tex] R_{10} = 2Ae^{-3/2}e^(r/a) [/tex] and [tex] Y_{00} = \frac{1}{\sqrt{4z\pi}} [/tex]

A is the so called variational parameter.

a)

Calculate the potential energy of the groundstate as a function of A.

b)

Calculate the kinetic energy of the groundstate as a function of A.

## Homework Equations

Schrödinger Equation:

[tex] -\frac{\hbar^2}{2m_e}\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2 \frac{\partial R}{\partial r}\right) +\frac{\hbar^2l(l+1)}{2m_er^2}R - \frac{Zr^2}{(4\pi\epsilon_0)r}R = ER [/tex]

## The Attempt at a Solution

I am not quite sure, but would I be ruight in saying that if a solve the LHS, and then divide through by R, this would give:

[tex] \frac{1}{R}\left( -\frac{\hbar^2}{2m_e}\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2 \frac{\partial R}{\partial r}\right) +\frac{\hbar^2l(l+1)}{2m_er^2}R - \frac{Zr^2}{(4\pi\epsilon_0)r}R\right) = E [/tex]

Would this be the potential energy, or is it the total energy?

Many thanks,

TFM