What is the difference between Quantum Mechanical Model of an atom and Bohr's atomic Model?
What is the difference between Quantum Mechanical Model of an atom and Bohr's atomic Model?
In the Bohr Model, the electron is treated as a particle in fixed orbits around the nucleus. In the Quantum Mechanical Model, the electron is treated mathematically as a wave. The electron has properties of both particles and waves.
The Bohr model was a one-dimensional model that used one quantum number to describe the distribution of electrons in the atom. The only information that was important was the size of the orbit, which was described by “n” the principle quantum number.
Schrodinger's model (Quantum Mechanical Model) allowed the electron to occupy three-dimensional space. It therefore required three coordinates, or three quantum numbers, to describe the distribution of electrons in the atom.
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the Bohr model treats the electron as a localized, classical object, not unlike a miniature billiard ball. But, in order to get a working model of the hydrogen atom using classical electrons, Bohr had to include a variety of rather strange, seemingly ad hoc rules. Among these are:
• The electron's angular momentum is always a positive integer multiple of Planck's reduced constant (quantization).
• Unlike macroscopic charged objects, electrons don't emit radiation as a result of their orbital motions (atomic stability).
• An electron transitioning between orbits emits or absorbs a photon (mechanism for emission and absorption of radiation).
This model works well for certain properties of hydrogen and hydrogen-like atoms, but breaks down when describing multi-electron atoms.
The contemporary model of the atom requires full blown quantum mechanics; an outgrowth of the de broglie hypothesis, which came 11 or maybe 12 years after Bohr's model.
Among the differences between the quantum mechanics description of the atom and Bohr's classical description, are:
• The quantum version makes predictions that are in agreement with experiment for all atoms; not just hydrogen-like ones.
• The quantum version predicts the observed fine structure spectral lines of hydrogen that Bohr's model does not.
• The quantum version correctly predicts the observed orbital angular momentum for the ground state electron in the hydrogen atom. The Bohr model does not.
• In the quantum version, the ad hoc rules Bohr used to explain observed behavior simply become artifacts of the deeper theory. They're no longer postulates plucked out of left field; they're consequences of something more fundamental.
There are probably lots of ways you can show the difference between the classical Bohr model and the quantum version. The one we'll use here is the difference in ground state orbital angular momenta predicted by each -- namely, we can show that the Bohr model predicts non-zero orbital angular momentum for the ground state electron in the hydrogen atom, whereas the quantum mechanics model predicts zero, in agreement with observation.
To see that Bohr's ground state has non-zero angular momentum, note that the classical formula for angular momentum is just mvrmvr, so that when we add in Bohr's quantization condition, above, we get the relation:
mvr=nℏmvr=nℏ
Since both nn and ℏℏ are greater than zero, the orbital angular momentum of the Bohr ground state electron is decidedly greater than zero.
To show that the quantum mechanics version has zero orbital angular momentum, we'll observe that QM tells us that the (squared) orbital angular momentum eigenvalue for a de Broglie electron, described by the wave function ψnlmψnlm, is just l(l+1)ℏ2l(l+1)ℏ2 (where l is called the "azimuthal" quantum number, for no damn good reason).
Plugging in the numbers we can see that the orbital angular momentum of the ground state wave function ψ100ψ100 is:
0(0+1)ℏ20(0+1)ℏ2
Which is indeed zero.
If you prefer to write out the math, you can derive (or just look up) the QM square orbital angular momentum operator, L2L2:
−ℏ2[1sinθ∂∂θ(sinθ∂∂θ)+1sin2θ∂2∂ϕ2]−ℏ2[1sinθ∂∂θ(sinθ∂∂θ)+1sin2θ∂2∂ϕ2]
and note (or look up) that the ground state eigenfunction for L2L2 is just some constant, so that the differential operator L2L2 produces the expected zero result.
So why does the de Broglie-like electron have zero orbital angular momentum in its ground state? Because it's in a spherically symmetric superposition of all possible orbital angular momentum states. And due to spherical symmetry, these all cancel out; they add to zero. Such behavior is only possible for de Broglie's wave-like electron; it is not possible, or even sensible, to imagine such behavior for a classical, billiard ball-like particle.
In the Bohr Model, the electron is treated as a particle in fixed orbits around the nucleus. In the Quantum Mechanical Model, the electron is treated mathematically as a wave. The electron has properties of both particles and waves.
The Bohr model was a one-dimensional model that used one quantum number to describe the distribution of electrons in the atom. The only information that was important was the size of the orbit, which was described by “n” the principle quantum number.
Schrodinger's model (Quantum Mechanical Model) allowed the electron to occupy three-dimensional space. It therefore required three coordinates, or three quantum numbers, to describe the distribution of electrons in the atom.
or
the Bohr model treats the electron as a localized, classical object, not unlike a miniature billiard ball. But, in order to get a working model of the hydrogen atom using classical electrons, Bohr had to include a variety of rather strange, seemingly ad hoc rules. Among these are:
• The electron's angular momentum is always a positive integer multiple of Planck's reduced constant (quantization).
• Unlike macroscopic charged objects, electrons don't emit radiation as a result of their orbital motions (atomic stability).
• An electron transitioning between orbits emits or absorbs a photon (mechanism for emission and absorption of radiation).
This model works well for certain properties of hydrogen and hydrogen-like atoms, but breaks down when describing multi-electron atoms.
The contemporary model of the atom requires full blown quantum mechanics; an outgrowth of the de broglie hypothesis, which came 11 or maybe 12 years after Bohr's model.
Among the differences between the quantum mechanics description of the atom and Bohr's classical description, are:
• The quantum version makes predictions that are in agreement with experiment for all atoms; not just hydrogen-like ones.
• The quantum version predicts the observed fine structure spectral lines of hydrogen that Bohr's model does not.
• The quantum version correctly predicts the observed orbital angular momentum for the ground state electron in the hydrogen atom. The Bohr model does not.
• In the quantum version, the ad hoc rules Bohr used to explain observed behavior simply become artifacts of the deeper theory. They're no longer postulates plucked out of left field; they're consequences of something more fundamental.
There are probably lots of ways you can show the difference between the classical Bohr model and the quantum version. The one we'll use here is the difference in ground state orbital angular momenta predicted by each -- namely, we can show that the Bohr model predicts non-zero orbital angular momentum for the ground state electron in the hydrogen atom, whereas the quantum mechanics model predicts zero, in agreement with observation.
To see that Bohr's ground state has non-zero angular momentum, note that the classical formula for angular momentum is just mvrmvr, so that when we add in Bohr's quantization condition, above, we get the relation:
mvr=nℏmvr=nℏ
Since both nn and ℏℏ are greater than zero, the orbital angular momentum of the Bohr ground state electron is decidedly greater than zero.
To show that the quantum mechanics version has zero orbital angular momentum, we'll observe that QM tells us that the (squared) orbital angular momentum eigenvalue for a de Broglie electron, described by the wave function ψnlmψnlm, is just l(l+1)ℏ2l(l+1)ℏ2 (where l is called the "azimuthal" quantum number, for no damn good reason).
Plugging in the numbers we can see that the orbital angular momentum of the ground state wave function ψ100ψ100 is:
0(0+1)ℏ20(0+1)ℏ2
Which is indeed zero.
If you prefer to write out the math, you can derive (or just look up) the QM square orbital angular momentum operator, L2L2:
−ℏ2[1sinθ∂∂θ(sinθ∂∂θ)+1sin2θ∂2∂ϕ2]−ℏ2[1sinθ∂∂θ(sinθ∂∂θ)+1sin2θ∂2∂ϕ2]
and note (or look up) that the ground state eigenfunction for L2L2 is just some constant, so that the differential operator L2L2 produces the expected zero result.
So why does the de Broglie-like electron have zero orbital angular momentum in its ground state? Because it's in a spherically symmetric superposition of all possible orbital angular momentum states. And due to spherical symmetry, these all cancel out; they add to zero. Such behavior is only possible for de Broglie's wave-like electron; it is not possible, or even sensible, to imagine such behavior for a classical, billiard ball-like particle.
