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Derivation of Schrodinger Wave Equation for Hydrogen Atom
Derivation of Schrödinger Wave Equation for Hydrogen Atoms
The Derivation of Schrödinger Wave Equation for Hydrogen Atoms should be investigated in detail.
Looking at the system of hydrogen atoms, one of the electrons moves around the nucleus. The nucleus is at the center, and the electron is subjected to the Coulomb force. According to classical mechanics, the motion of the electron is constrained by the potential energy, whose potential energy is\ (V = -\ frac {e ^ {2}} {4\ pi\ epsilon_ {0} r}\), where\ (e\) is the electron charge,\ (\ epsilon_ {0}\) is the vacuum dielectric constant, and\ (r\) is the distance between the electron and the nucleus.
In quantum mechanics, the wave function\ (\ Psi\) describes the state of microscopic particles. The form of the Schrodinger equation is\ (H\ Psi = E\ Psi\),\ (H\) is the Hamiltonian operator, and\ (E\) is the system energy.
In the case of hydrogen atoms, the Hamiltonian operator\ (H = -\ frac {\ hbar ^ {2}} {2m}\ nabla ^ {2} + V\),\ (\ hbar\) is the reduced Planck constant,\ (m\) is the electron mass, and\ (\ nabla ^ {2}\) is the Laplace operator.
Substituting the potential energy\ (V = -\ frac {e ^ {2}} {4\ pi\ epsilon_ {0} r}\) into the Hamiltonian gives\ (H = -\ frac {\ hbar ^ {2}} {2m}\ nabla ^ {2} -\ frac {e ^ {2}} {4\ pi\ epsilon_ {0} r}\).
Then the Schrodinger equation for the hydrogen atom is\ (\ left (-\ frac {\ hbar ^ {2}} {2m}\ nabla ^ {2} -\ frac {e ^ {2}} {4\ pi\ epsilon_ {0} r}\ right) \ Psi = E\ Psi\).
To solve this equation, a spherical coordinate system is often used. In the spherical coordinate system, the Laplacian operator\ (\ nabla ^ {2} =\ frac {1} {r ^ {2}}\ frac {\ partial} {\ partial r}\ left (r ^ {2}\ frac {\ partial} {\ partial r}\ right) +\ frac {1} {r ^ {2}\ sin\ theta}\ frac {\ partial} {\ partial\ theta}\ left (\ sin\ theta\ frac {\ partial} {\ partial\ theta}\ right) +\ frac {1} {r ^ {2}\ sin ^ {2}\ theta}\ frac {\ partial ^ {2}} {\ partial\ varphi ^ {2}} \).
Substitute it into the Schrodinger equation, separate the variables, and let\ (\ Psi (r,\ theta,\ varphi) = R (r) Y (\ theta,\ varphi) \), solve the radial equation and the angular equation respectively.
The radial distribution of electrons can be obtained by solving the radial equation, and the angular momentum quantization can be obtained by solving the angular equation. Through a series of mathematical operations and derivations, the solution of the Schrödinger wave equation of the hydrogen atom can be obtained. This solution can describe various states of electrons in the hydrogen atom, such as energy and angular momentum. It is consistent with the experimental results and lays a foundation for the quantum mechanical description of the atomic structure.
The Derivation of Schrödinger Wave Equation for Hydrogen Atoms should be investigated in detail.
Looking at the system of hydrogen atoms, one of the electrons moves around the nucleus. The nucleus is at the center, and the electron is subjected to the Coulomb force. According to classical mechanics, the motion of the electron is constrained by the potential energy, whose potential energy is\ (V = -\ frac {e ^ {2}} {4\ pi\ epsilon_ {0} r}\), where\ (e\) is the electron charge,\ (\ epsilon_ {0}\) is the vacuum dielectric constant, and\ (r\) is the distance between the electron and the nucleus.
In quantum mechanics, the wave function\ (\ Psi\) describes the state of microscopic particles. The form of the Schrodinger equation is\ (H\ Psi = E\ Psi\),\ (H\) is the Hamiltonian operator, and\ (E\) is the system energy.
In the case of hydrogen atoms, the Hamiltonian operator\ (H = -\ frac {\ hbar ^ {2}} {2m}\ nabla ^ {2} + V\),\ (\ hbar\) is the reduced Planck constant,\ (m\) is the electron mass, and\ (\ nabla ^ {2}\) is the Laplace operator.
Substituting the potential energy\ (V = -\ frac {e ^ {2}} {4\ pi\ epsilon_ {0} r}\) into the Hamiltonian gives\ (H = -\ frac {\ hbar ^ {2}} {2m}\ nabla ^ {2} -\ frac {e ^ {2}} {4\ pi\ epsilon_ {0} r}\).
Then the Schrodinger equation for the hydrogen atom is\ (\ left (-\ frac {\ hbar ^ {2}} {2m}\ nabla ^ {2} -\ frac {e ^ {2}} {4\ pi\ epsilon_ {0} r}\ right) \ Psi = E\ Psi\).
To solve this equation, a spherical coordinate system is often used. In the spherical coordinate system, the Laplacian operator\ (\ nabla ^ {2} =\ frac {1} {r ^ {2}}\ frac {\ partial} {\ partial r}\ left (r ^ {2}\ frac {\ partial} {\ partial r}\ right) +\ frac {1} {r ^ {2}\ sin\ theta}\ frac {\ partial} {\ partial\ theta}\ left (\ sin\ theta\ frac {\ partial} {\ partial\ theta}\ right) +\ frac {1} {r ^ {2}\ sin ^ {2}\ theta}\ frac {\ partial ^ {2}} {\ partial\ varphi ^ {2}} \).
Substitute it into the Schrodinger equation, separate the variables, and let\ (\ Psi (r,\ theta,\ varphi) = R (r) Y (\ theta,\ varphi) \), solve the radial equation and the angular equation respectively.
The radial distribution of electrons can be obtained by solving the radial equation, and the angular momentum quantization can be obtained by solving the angular equation. Through a series of mathematical operations and derivations, the solution of the Schrödinger wave equation of the hydrogen atom can be obtained. This solution can describe various states of electrons in the hydrogen atom, such as energy and angular momentum. It is consistent with the experimental results and lays a foundation for the quantum mechanical description of the atomic structure.
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