# Question About Bohr's Quantified Shell Model

1. Nov 2, 2011

### octagon

I'm sure this has a straightforward explanation and am hoping someone can answer it for me.

Looking at this diagram of Bohr's quantified shell model of the atom:

http://library.thinkquest.org/C005775/Theory/oldtheory_section3.html

...I don't understand what is preventing the electron from colliding with the nucleus after n=1, since it still has one unit of energy left, and each previous orbit only required one unit of energy to move it to the next smaller orbit.

Thanks.

Last edited by a moderator: Apr 26, 2017
2. Nov 2, 2011

### Bill_K

octagon, The Bohr theory was an early attempt to explain the atom in nearly classical terms. Electrons were supposed to be like little BBs that circled the nucleus in well-defined orbits, and emitted radiation by jumping from one orbit to another. It was soon realized that while this theory explained some features, it was quite wrong on others. It was replaced by the Schrodinger theory in 1926. The Bohr theory is now of historical interest only. On the web page you're reading, if you go ahead a few steps you'll find this explained.

3. Nov 2, 2011

### ytuab

The total enegies are the same in both Bohr model and Schrodinger's hydrogen.
As both satisfy the Virial theorem, 2K = - V, the average kinetic energies (K) and potential energies (V) are the same, too.

In n=1 state (1s), Bohr model electron doesn't collide with nucleus, because the angular momentum is not zero.
To get the kinetic energies the same, Schrodinger's hydrogen needs to have "radial" kinetic energy, because the angular momentum is zero ( L = 0 ) in 1s state.

When we solve the Schrodinger equation for hydrogen, we choose the next boundary condition to prevent divergence,

$$u_l (r) = rR_l (r) \qquad u_l (r) \to e^{-\rho r} \quad ( r \to \infty) \qquad u_l (r) \to r^{l+1} \quad ( r \to 0 )$$
The instant you choose this boundary condition, a "mountain" (= wavelength) is generated in the "radial" direction
This means "radial" kinetic enegy is not zero, when you choose this boundary condition.
Instead you have to get the angular momentum (= tangential kinetic energy) zero ( L=0 ) in the Schrodinger equation.

By the way, when the angular momentum is not zero, it is impossible that the electron comes closer to the nucleus than a point (perihelion) in Bohr's planetary model.
When the angular momentum is L (of course, L is constant), the "tangential" kinetic energy (T) is,

$$mv_{\theta} r = L \quad \to \quad T = \frac{1}{2} mv_{\theta}^2 =\frac{1}{2} \frac{L^2}{m r^2}$$
So this tangential kinetic energy increases at the inverse square of the radius (1/r^2).
At the points near r = 0, the sum of tangential kinetic energy and potential energy is

$$T + V = \frac{1}{2} \frac{L^2}{m r^2} - \frac{e^2}{4\pi\epsilon r} \to +\infty \quad ( r \to 0 )$$
So to keep the total energy ( E < 0 ) constant, the "radial" kinetic energy (R) becomes minus in Schrodinger equation within perihelion.

$$R = \frac{1}{2}m v_r^2 < 0 \qquad T = \frac{1}{2} m v_{\theta}^2 > 0 \qquad ( R + T + V = E < 0 )$$
This result shows that "minus" electron mass ( m < 0 ) and "plus" electron mass ( m > 0) are mixed in Schrodinger equation ?