Top page (correct Bohr model including the two-electron atoms)

Electron spin is an illusion !

In this top page, we calculate the reduced mass of the helium atom, which belongs to the *three-body system*.

And we use this reduced mass except when the center of mass is at the helium nucleus.

This method can give *more accurate* result than the quantum-mechanical variational method.

*(Fig.1)*

First, we explain the reduced mass of the two-body hydrogen atom.

Because the reduced mass of the helium atom is based on this hydrogen.

We suppose the particle 1 is a hydrogen nucleus, and the particle 2 is an electron.

We define as follows,

*(Eq.1)*

where r_{1} and r_{2} mean the *position vectors* of each particle.

When the *force* between the two particles is **f(r)**, the *equation of motion* in the particle 1 is

*(Eq.2)*

where m_{1} means the mass of the particle 1.

And the particle 2 satisfies

*(Eq.3)*

These relations satisfy the **law of action and reaction**.

Subtracting Eq.2 from Eq.3, we obtain

*(Eq.4)*

where Eq.1 is used.

If we use the reduced mass (μ) of

*(Eq.5)*

Eq.4 can be expressed as

*(Eq.6)*

Eq.6 means that we need to use the reduced mass (μ), if we consider that the nucleus is at rest.

For the reduced mass to be effective, we need to confirm the nuclear kinetic energy is **absorbed** into another light particle with reduced mass.

The coordinate of the center of mass (R) is

*(Eq.7)*

From Eq.1 and Eq.7,

*(Eq.8)*

In the same way, we obtain

*(Eq.9)*

So the total kinetic energy of the two particles are

*(Eq.10)*

As shown in Eq.10, the total kinetic energy is the sum of the movement of the center of mass and the relative motion of the particles.

From Eq.6 and Eq.10 if we use the reduced mass, we can treat the nucleus as if it is at rest inside atoms with respect to the force and energy.

*(Eq.11) Relation between redued mass and velocities.*

In Eq.10, when the total velocity of the center of mass (= R ) is zero, the total kinetic energy is expressed **only** by reduced mass (= μ ) and relative velocity (= dot r ).

( In Eq.11, total momentum is zero, because m_{2} and m_{1} are moving in the **opposite** directions. )

It seems that total momentum is not zero in Eq.11 below, because **only** mass μ is moving in one direction.

In fact, the mass of heavy m_{2} automatically becomes **infinite**, when the reduced mass μ is used.

So as this infinite mass × velocity_{2} (= 0 ) is **not** zero, total momentum including μ and m_{2} is conserved.

In the case of the three-body problem like the helium, it is a little complicated.

See also this page.

*(Fig.2) Hydrogen and helium nuclei.*

In case of hydrogen (= Fig.2 left ), nucleus is **always** moving due to two-body relation.

But in case of Fig.2 right ( two electrons are just opposite of nucleus ), its helium nucleus is **stationary**, because it's just in **equilibrium** between two electrons.

Of course, it is known that Fig.2 right helium model gives a little different value from actual ground state energy.

So how about the case when two electron's orbits are just **perpendicular** to each other ?

( Quantum mechanical **vague** wavefunction **cannot** distinguish these differences. This is also one of the reasons why quantum mechanics is **wrong**. )

*(Fig.3) Nucleus is "oscillating" instead of rotating. (= special case)*

First we think about the very **special** case, in which the nucleus is **oscillating** like a spring. (= **straight-line motion**, Fig.3. )

Different from Fig.8 (= normal pattern ), the nuclear ( linear ) velocity becomes **maximum**, when the center of mass is at the nucleus ( Fig.3 middle ).

And in the upper (= A. ) and lower (= C.) panel of Fig.3, the nucleus **turns back**, so its velocity becomes **zero** at these points.

sample JAVA program (= the nucleus is oscillating like Fig.3)

C language program

If you copy and paste the above program source code into a text editor, you can easily compile and run this.

In this program, we first input the initial **y-coordinate r2** (in MM) of electron 1 (see Fig.4), and the absolute value of the total energy E (in eV) of Helium.

From the inputted values, this program outputs the x component of electron 1 velocity in Fig.5, and WN (= 1/4 de Broglie's waves ). Here 1 SS = 1 × 10^{-23} second.

*(Fig.4) Starting positions of Fig.3 model.*

*(Fig.5) Electrons have moved one quarter of their orbits.*

We choose the positions of Fig.4 as the **starting points**.

Because in Fig,3 moving pattern, the helium nucleus **stops** when the electrons are at the positions of Fig.3 A and C.

So we use the **usual electron mass** (NOT reduced mass), when we calculate the initial velocity of Fig.4.

(After starting, we use the reduced mass.)

Table 1 shows the result of "**nuclear oscillating model**" of Fig.3. ( last VX = 0 )

E (eV) | r2 (MM) | WN | WN x 4 |
---|---|---|---|

-79.0000 | 3137.0 | 0.250011 | 1.000044 |

-79.0037 | 3137.5 | 0.250005 | 1.000020 |

-79.0069 | 3136.5 | 0.250000 | 1.000000 |

-79.0100 | 3136.0 | 0.249996 | 0.999984 |

-79.0300 | 3135.5 | 0.249964 | 0.999856 |

As shown in Table 1, r2 is a little longer than r1 due to the repulsive force between electrons.

And the result of Table 1 is **-79.0069 eV**, which is a little different from **-79.0037 eV** of top page.

( Experimental value is **-79.005 eV**. )

But of course, this result is more accurate than latest quantum mechanical method (= **-79.015 eV** ).

Only the relativistic correction can not be explained in Fig.3 (= Table 1) model.

The nuclear linear motion in Fig.3 model is very "**special**" case, which doesn't occur naturally.

For example, the planets which attract each other **usually rotate** (not oscillate).

*(Fig.6) Starting point = NOT reduced mass= Virtual particle stops.*

Unlike the two-body hydrogen atom, the oscillating motion of Fig.3 includes two points at which the **nucleus stops** (= A and C of Fig.3 ).

At these points of A and C of Fig.3, the **virtual particle between the two electrons stops**, too.

So if we choose A of Fig.3 as the starting point of the program, we should not use the reduced mass in calculating the initial electron's velocity.

Because using reduced mass means that the virtual particle and nucleus are behaving like hydrogen atom.

*(Fig.7) Reduced mass of Helium atom.*

At other points, the nucleus and virtual particle are moving (linearly) in Fig.3.

And the virtual particle is a single existence influencing the nuclear motion.

So in this case, we can use reduced mass in other moving positions.

The electron number is "2" in helium, so each electron is a half of the overall reduced mass,

*(Eq.12)*

This correct discrimination between reduced mass and usual mass gives real and **exact** results of the helium ground state energy.

*(Fig.8) Nucleus and virtual particle are "rotating".*

Actually, when the virtual particle between two electrons is at the nucleus (= **center of mass** ), they **stop** temporarily (= Fig.8B, Fig.10 ).

Because Fig.8B and Fig.10 (= two electrons are at just opposite sides of their nucleus ) are most **stable** and "equilibrium" point, considering Coulomb **repulsive** force.

*(Fig.9) Only "tangential" velocity is left at center of mass.*

Considering the **conservation** of angular momentum, **tangential** ( rotating ) velocity of the nucleus becomes much **bigger** than the linear motion at Fig.8B, when it is **closer** to center of mass.

( Imagine rotating figure skater. )

So we can **neglect** linear motion of nucleus, when the center of mass is at origin.

Of course, when the rotating radius is **zero**, tangential velocity of nucleus also becomes close to **zero**.

So if we choose Fig.8 model, we should use **usual mass** ( not reduced mass ), when two electrons are just at opposite sides of nucleus.

*(Fig.10) Most stable positions = NOT reduced mass. *

In Fig.10, two electrons are just at the opposite sides of the nucleus.

But before and after that, due to the de Broglie's **interference**, two orbits have to be just perpendicular to each other.

( If they are not perpendicular, the opposite wave phases of them cancel each other out and let them unstable. )

This force is caused by the de Broglie wave's interference, and of course, due to the **law of action and reaction**, the nucleus is **pushed** in the opposite direction inside the helium.

The point is that when the center of the two electrons are just **equal** to the nucleus, they become **most stable**.

This is like the **bottom** of the curve.

So only at this **equilibrium** point (= Fig.10 ), the nucleus **stops**, and we have to use the usual mass in calculating the initial electron's velocity.

*(Fig.11) Two electrons are moving (= reduced mass ).*

The instant two electrons start to move in the direction of Fig.11, their nucleus **goes away** from the center of mass.

So to stop the nucleus, we have to use reduced mass, as I explain in Eq.11

E (eV) | r1 (MM) | WN | WN x 4 |
---|---|---|---|

-78.9900 | 3075.0 | 0.250022 | 1.000088 |

-79.0000 | 3074.0 | 0.250006 | 1.000024 |

-79.0037 | 3074.0 | 0.250000 | 1.000000 |

-79.0100 | 3074.0 | 0.249990 | 0.999960 |

-79.0300 | 3073.0 | 0.249958 | 0.999832 |

This result (= **-79.0037 eV** ) is just **consistent** with the experimental results ( except for the relativistic effect ).

This result clearly shows Fig.8-Fig.11 helium model is most **accurate**.

We use the following program. See also this page in detail.

In this program, we can **choose** the usual mass or reduced mass in the electron.

sample JAVA program 1 SS = 1 × 10^{-23}

C language program

Table 3 shows the "**Not reduced mass** condition" results.

E (eV) | r1 (MM) | WN | WN x 4 |
---|---|---|---|

-79.0037 | 3075.0 | 0.250071 | 1.000284 |

-79.0300 | 3074.0 | 0.250030 | 1.000120 |

-79.0485 | 3073.5 | 0.250000 | 1.000000 |

-79.0600 | 3073.0 | 0.249982 | 0.999928 |

-79.0800 | 3072.0 | 0.249950 | 0.999800 |

As shown in Table 3, if we don't use the reduced mass in the neutral helium atom, the calculation result becomes **-79.0485 eV**, which is a little different from the experimental value (-79.005147 eV).

(The error is -79.0485 - (-79.005147) = **-0.04335 eV**. Compare this with Table 2 result of **0.001447 eV**.)

Of course, this result is also very close to the experimental value.

But we have to use the reduced mass to get most **correct** ground state energy in helium.

Table 4 shows the "**All reduced mass** condition" results.

E (eV) | r1 (MM) | WN | WN x 4 |
---|---|---|---|

-79.0037 | 3075.0 | 0.250037 | 1.000148 |

-79.0200 | 3074.5 | 0.250011 | 1.000044 |

-79.0268 | 3074.0 | 0.250000 | 1.000000 |

-79.0300 | 3074.0 | 0.249995 | 0.999980 |

-79.0500 | 3073.0 | 0.249963 | 0.999852 |

In Table 4, we **always** use reduced mass, even when the center of two electrons is just equal to the nucleus.

This result is also a little different from the experimental result.

( Though it is almost the same. )

So we have to **discriminate** "reduced" or "usual" mass, depending on their positional relation, to get most accurate value.

*(Fig.12) Electron pushes nucleus (= law of action and reaction ). *

As I said above, when two electrons move in the orbitals **perpendicular** to each other, they are influenced by de Broglie wave's **interference**.

Of course, when the center of these two electrons goes away from the center of mass, helium nucleus are **pushed** by electrons due to law of action and reaction.

In Fig.12, only the direction of two particles' crashing is important (= v, Fig.12 lower ).

( The parallel component doesn't change. )

*(Eq.13) Total energy conservation.*

As shown in Eq.13, total ( kinetic ) energies are conserved before and after this collision.

In Eq.13, "m" is electron's mass, "M" is nuclear mass.

*(Eq.14) Total momentum.*

And total momentum is conserved, too.

From Eq.13 and Eq.14, we get each velocity of

*(Eq.15) Each velocity.*

You can confirm this result by substituting Eq.15 into Eq.13 and Eq.14.

As shown in Eq.15, the **relative** velocity of Eq.11 is almost conserved.

*(Eq.16)*

So if we get initial electron's velocity (= v ) using **usual** mass in Fig.10, that velocity can be used as it is, even in **relative** coordinate.

*(Eq.17)*

But if you use the initial velocity gotten by reduced mass (= μ ), this initial velocty becomes **faster** on the condition of the **same** kinetic energy K.

Because the reduced mass is **smaller** than usual mass.

So if you suddenly change initial velocity into one gotten by reduced mass, it means this electron suddenly becomes **faster** !

This is strange.

So the initial velocity by usual mass remains the same even in relative coordinate.

Total momentum is always conserved, even when only the initial nuclear velocity is **zero**.

Because as shown in Eq.11, nuclear mass becomes **infinite**, the moment we start to use reduced mass.

2013/9/30 updated. Feel free to link to this site.