Saturday, February 7, 2009

Schrodinger’s Equation

Quantum mechanics describes the spectra in a much better way than Bohr’s model.

Electron has a wave character as well as a particle character. The wave function of the electron ψ(r,t ) is obtained by solving Schrodinger’s wave equation. The probability of finding an electron is high where | ψ(r,t )|² is greater. Not only the information about the electron’s position but information about all the properties including energy etc. that we calculated using the Bohr’s postulates are contained in the wave function of ψ(r,t).

Quantum Mechanics of the Hydrogen Atom

The wave function of the electron ψ(r,t) is obtained from the Schrodinger’s equation

-(h²/8π²m) [∂²ψ /∂x² + ∂²ψ /∂y² + ∂²ψ/∂z²] - Ze²ψ/4πε0r = E ψ


where
(x.y,z ) refers to a point with the nucleus as the origin and r is the distance of this point from the nucleus.
E refers to the energy.
Z is the number of protons.

There are infinite number of functions ψ(r,t) which satisfy the equations.

These functions may be characterized by three parameters n,l, and ml.

For each combination of n,l, and ml there is an associated unique value of E of the atom of the ion.

The energy of the wave function of characterized by n,l, and ml depends only on n and may be written as


En = - mZ²e4/8 ε0²h²n²

These energies are identical with Bohr’s model energies.

The paramer n is called the principal quantum number, l the orbital angular momentum quantum number and ml. The magnetic quantum number.

When n = 1, the wave function of the hydrogen atom is

ψ(r) = ψ100 = √(Z³/ π a0²) *(e-r/ a0)


ψ100 denotes that n =1, l = 0 and ml = 0

a0 = Bohr radius

In quantum mechanics, the idea of orbit is invalid. At any instant the wve function is spread over large distances in space, and wherever ψ≠ 0, the presence of electron may be felt.

The probability of finding the electron in a small volume dV is | ψ(r)| ² dV
We can calculate the probability p(r)dr of finding the electron at a distance between r and r+dr from the nucleus.

In the ground state for hydrogen atom it comes out to be

P(r) = (4/ a0)r²e -2r/ a0

The plot of P(r) versus r shows that P(r) is maximum at r = a0 Which the Bohr’s radius.

But when we put n =2, the maximum probability comes at two radii one near r = a0 and the other at r = 5.4 a0. According to Bohr model all electrons should be at r = 4 a0.

3 comments:

Prudhvi Raj Borra said...

I don't know who has provided this.But this is simply awesome.Ur explanation was excellent.Plz provide us with some other important matters in quantum mechanics so that we can be able to know the maximum bits in IIT.Plz send me some materials which are necessary for IIT.My email address is prudhvirajborra@yahoomail.com.Plz send me.Thank you.

Prudhvi Raj Borra said...

Thanks for this.Publish more information on quantum mechanics.Post me some matter to my email so that i can secure a seat in iit.Plz send me some material which is essential for iit.Plz.Thank you.

KVSSNrao said...

This explanation is posted in response to question put by a member of the IIT JEE Academy orkut community. Please mention the specific topic on which you need more explanation. We can spend some time on it. I post every thing on these blogs. There is nothing extra for sending by mail.

Thank you for the appreciation.