Today in class we discussed an "incorrect atomic model" situation worded like this:
AN electron is located inside the nucleus of an atom. For H-atoms, the electron is confined to its nuclear diameter, 1.7*10^-15 m. The process to get to the answer, which was 3.4*10^10 m/s, was very confusing for me. Can someone please re-explain this model he used in class and also explain how it exactly pertains to the Heisenberg Uncertainty Equation?
Incorrect Atomic Model
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Re: Incorrect Atomic Model
The first thing to note is that he showed us this example to demonstrate that the Heisenberg equation can be used to prove some models impossible.
The example gives you the diameter of the atom, which is 1.7 * 10^-15 and is looking for the uncertainty of the electron's uncertainty of position, dv. He gave us two equations to work with which were "dp= m*dv" and "dp*dx is greater or equal to h/4pi". The dv is in dp=m*dv equation but all we have to work with so far is the nuclear diameter or dx. So we first use the "dp*dx is greater or equal to h/4pi" so we can find the dp value to plug back into the "dp=m*dv". We can plug in planck's constant(6.626*10^-34) and dx(1.7*10^-15) into the "dp*dx is greater or equal to h/4pi" equation and it will give us back dp=3.1*10^-20 kg*m*s^-1. Plug that into dp=m*dv, along with the mass of an electron(9.1*10^-31) and solve for dv. Dv will equal 3.4*10^10 m*s^-1.
The important thing to understand is that just from the uncertainty, 3.4*10^10, is larger than the speed of light, 3.0*10^8. This is impossible, demonstrating that it is not possible for the electron to be in the nucleus of the atom. The atom has a physical limit to how small it could be shown in factors like mass of electrons and speed of light. Hope this helps!
The example gives you the diameter of the atom, which is 1.7 * 10^-15 and is looking for the uncertainty of the electron's uncertainty of position, dv. He gave us two equations to work with which were "dp= m*dv" and "dp*dx is greater or equal to h/4pi". The dv is in dp=m*dv equation but all we have to work with so far is the nuclear diameter or dx. So we first use the "dp*dx is greater or equal to h/4pi" so we can find the dp value to plug back into the "dp=m*dv". We can plug in planck's constant(6.626*10^-34) and dx(1.7*10^-15) into the "dp*dx is greater or equal to h/4pi" equation and it will give us back dp=3.1*10^-20 kg*m*s^-1. Plug that into dp=m*dv, along with the mass of an electron(9.1*10^-31) and solve for dv. Dv will equal 3.4*10^10 m*s^-1.
The important thing to understand is that just from the uncertainty, 3.4*10^10, is larger than the speed of light, 3.0*10^8. This is impossible, demonstrating that it is not possible for the electron to be in the nucleus of the atom. The atom has a physical limit to how small it could be shown in factors like mass of electrons and speed of light. Hope this helps!
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Re: Incorrect Atomic Model
Essentially we are trying to use the Heisenberg Uncertainty Model to disprove the atomic model that an electron is located inside the nucleus of an atom with diameter 1.7*10^-15 m (electrons are located in the space around the nucleus). We used the uncertainty model to determine the delta velocity (uncertainty) of the electron, which turned out to be 3.4*10^10 m/s. Since that value is greater than the speed of light, making it impossible, we can conclude that the atomic model given is incorrect by using the Heisenberg Uncertainty Equation.
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Re: Incorrect Atomic Model
We used the Heisenberg Indeterminacy Equation (delta p)(delta x) > or equal to h/4Pi to find our final answer of 3.4 x 10^10 m/s. Our delta x was 1.7 x 10^-15 m and instead of solving for the momentum right away, we had to find the velocity, since the only other information that we know is the mass of an electron and momentum = (mass)(velocity). Since the velocity is greater than the speed of light, we proved that the electron cannot be located in the nucleus.
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Re: Incorrect Atomic Model
If delta v signifies a range of possible velocities of an electron, isn't it possible that the velocity could be less than the speed of light? How do we know that the velocity of the electron is more than the speed of light if the range doesn't specify whether of not it is?
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Re: Incorrect Atomic Model
Delta v is the amount of uncertainty in the electron's velocity. Because the uncertainty was greater than the speed of light and was an unrealistic number, it proves that electrons do not collapse to the nucleus.
Re: Incorrect Atomic Model
With the given delta x, which is the diameter of the atom, we can calculate the minimum delta momentum with the Heisenberg uncertainty principal (the equation). Because the minimum velocity calculated then is larger than the velocity of light, which should be the largest velocity possible, the model is incorrect.
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