Write solution of Schrodinger's Equation for H-atom.
\[ H \psi(\mathbf{r}, t) = E \psi(\mathbf{r}, t) \]
Where:
- \( H \) is the Hamiltonian operator, representing the total energy of the system.
- \( \psi(\mathbf{r}, t) \) is the wave function of the electron, which depends on the position vector \( \mathbf{r} \) and time \( t \).
- \( E \) is the energy of the system.
The Hamiltonian operator for a single electron in a hydrogen atom is:
\[ H = -\frac{\hbar^2}{2m_e} \nabla^2 - \frac{e^2}{4\pi\epsilon_0 r} \]
Where:
- \( \hbar \) is the reduced Planck's constant (\( \hbar = \frac{h}{2\pi} \)).
- \( m_e \) is the mass of the electron.
- \( \nabla^2 \) is the Laplacian operator (the sum of the second partial derivatives with respect to \( x, y, \) and \( z \)).
- \( e \) is the elementary charge (charge of the electron).
- \( \epsilon_0 \) is the vacuum permittivity (electric constant).
- \( r \) is the distance between the electron and the nucleus (the position vector magnitude).
Now, let's solve the Schrödinger's equation for the hydrogen atom. To do this, we assume the wave function can be separated into radial and angular parts as follows:
\[ \psi(\mathbf{r}, t) = R(r) Y(\theta, \phi) \]
Where:
- \( R(r) \) is the radial part, dependent only on \( r \).
- \( Y(\theta, \phi) \) is the angular part, dependent only on the angles \( \theta \) and \( \phi \).
Substituting this into the Schrödinger's equation and rearranging, we get:
\[ \frac{1}{R(r)} \left(-\frac{\hbar^2}{2m_e} \frac{d^2 R(r)}{dr^2} - \frac{\hbar^2}{2m_e} \frac{2}{r} \frac{dR(r)}{dr} \right) + \frac{2m_e e^2}{4\pi\epsilon_0 \hbar^2 r} R(r) = E \]
To simplify this equation, we use the separation constant \( \lambda \) such that:
\[ \frac{1}{R(r)} \left(-\frac{\hbar^2}{2m_e} \frac{d^2 R(r)}{dr^2} - \frac{\hbar^2}{2m_e} \frac{2}{r} \frac{dR(r)}{dr} \right) = \lambda \]
This leads to two differential equations: one for \( R(r) \) and the other for \( Y(\theta, \phi) \).
Solving the radial equation yields:
\[ -\frac{\hbar^2}{2m_e} \frac{d^2 R(r)}{dr^2} - \frac{\hbar^2}{2m_e} \frac{2}{r} \frac{dR(r)}{dr} + \frac{2m_e e^2}{4\pi\epsilon_0 \hbar^2 r} R(r) = \lambda R(r) \]
To solve this, we use the change of variables \( \rho = 2r/\alpha \), where \( \alpha \) is the Bohr radius (\( \alpha = \frac{4\pi\epsilon_0\hbar^2}{m_e e^2} \)).
After simplification, we get the radial wave function \( R(r) \) as:
\[ R(\rho) = N \rho^l e^{-\rho/2} L_{n-l-1}^{2l+1}(\rho) \]
Where:
- \( N \) is a normalization constant.
- \( n \) is the principal quantum number (integer \(\geq 1\)).
- \( l \) is the azimuthal quantum number (integer \(\geq 0\) and \(< n\)).
- \( L_{n-l-1}^{2l+1}(\rho) \) is the associated Laguerre polynomial.
For the angular part, the solution is given by the spherical harmonics \( Y(\theta, \phi) \).
Therefore, the complete wave function for the hydrogen atom is given by:
\[ \psi_{n, l, m}(\mathbf{r}, t) = R_{n, l}(r) Y_{l, m}(\theta, \phi) e^{-i E_{n}t/\hbar} \]
Where:
- \( n, l, m \) are the quantum numbers (with \( n \) and \( l \) defined as above, and \( m \) is the magnetic quantum number).
- \( R_{n, l}(r) \) is the radial part.
- \( Y_{l, m}(\theta, \phi) \) is the spherical harmonic.
- \( E_{n} \) is the energy associated with the state defined by the quantum numbers \( n, l \) (given by \( E_{n} = -\frac{m_e e^4}{32\pi^2\epsilon_0^2\hbar^2 n^2} \)).
In summary, the wave function of the hydrogen atom is a product of the radial part and the spherical harmonic, and it depends on the principal quantum number \( n \), the azimuthal quantum number \( l \), and the magnetic quantum number \( m \). The energy of the system is determined by the principal quantum number \( n \).
No comments: