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Schrodinger Equation (Hydrogen)

This is possibly a question for the physics forum, but the bulk of the meat is maths, despite this question being an extension in my physical chemitry textbook! Here goes...

The form of the Schrodinger equation for the energy E of the 1s orbital in the hydrogen atom is as follows:

-\frac{h^2}{8\pi^2 m} [\frac{d^2 \Psi}{dr^2} + (\frac{2}{r}) \frac{d \Psi}{dr}] - \frac{e^2}{4 \pi \epsilon_0 r}\Psi = E\Psi

where

h

is the planck constant,

m

is the mass of the electron,

e

is the charge on the electron and

\epsilon_0

is the fundamental constant; vaccum permittity. The second (square bracketed) term is the kinetic energy of the electron, and the third term is potential energy due to electrostatic attraction to a single proton at a distance r. Also note that

a_0=\frac{\epsilon_0 h^2}{\pi m e^2}

.

Find

E

.

My first instinct was to multiply both sides by

\frac{-8\pi^2 m}{h^2}

and collect together fundamental constants using

a_0=\frac{\epsilon_0 h^2}{\pi m e^2}

, which gets me to

\frac{d^2 \Psi}{dr^2} +(\frac{2}{r}) \frac{d \Psi}{dr} + \frac{2}{a_0 r} \Psi = (\frac{-8 \pi^2 m E}{h^2}) \Psi

,

which is a little hairy.

Can anyone point me in the right direction?

Edit: If you plug in all of the constants into "E=..." and multiply by the Avogrado constant, you get

1310 KJ mol^{-1}

, the first ionisation enthalphy for hydrogen; pretty cool.

(edited 12 years ago)

Reply 1

SimonM

It's a differential equation. No physics here at all - just solve it.

Reply 2

DFranklin

The first two terms (the ones with the 1st and 2nd derivatives of phi) can be written as

r^m \dfrac{d}{dr}\left( r^n \dfrac{d\Phi}{dr}\right)

for suitable choices of m and n.

Reply 3

Perpetuallity

Original post by SimonM

It's a differential equation. No physics here at all - just solve it.

Original post by DFranklin

The first two terms (the ones with the 1st and 2nd derivatives of phi) can be written as

r^m \dfrac{d}{dr}\left( r^n \dfrac{d\Phi}{dr}\right)

for suitable choices of m and n.

Right. I'm trying to avoid solving this differential equation by direct methods. It looks nasty. After some google-ing, it turns out that the second and third terms cancel out if

\frac{d \Psi}{dr} = \frac{-\Psi}{a_0}

which is the equation for exponential decay, whose solution is

\Psi = \Psi_0 e^{-r/a_0}

The energy for the 1s orbital is found by differentiating my solution attempt in my OP again, apparently (instinct got me somewhere at least)

\Rightarrow \frac{d^2 \Psi}{dr^2} = \frac{d}{dr}(\frac{d \Psi}{dr}) = \frac{d}{dr}(\frac{\Psi}{a_0}) = \frac{\Psi}{{a_0}^2}

\frac{1}{{a_0}^2} = \frac{-8 \pi^2 mE}{h^2}

\therefore E=\frac{-h^2}{8 \pi^2 m{a_0}^2}

Pretty cool. Cheers.

Edit: Actually, now with the benefit of heinsight, the DE doesn't look too bad.

(edited 12 years ago)

Reply 4

soutioirsim

Original post by Perpetuallity

\Rightarrow \frac{d^2 \Psi}{dr^2} = \frac{d}{dr}(\frac{d \Psi}{dr}) = \frac{d}{dr}(\frac{\Psi}{a_0}) = \frac{\Psi}{{a_0}^2}

Sorry I'm just being nosy here but could you explain to me how you got the last part of this line? (I can't see how you've differentiated that :colondollar: