Spherically symmetric states in the hydrogen atom

In summary, the equation given is the Schrodinger equation for spherically symmetric functions in a hydrogen-like atom. By substituting an assumed solution, the values of b and the ratio B/A can be found. This solution corresponds to the second energy level and has a coefficient of b equal to -Z/2a0. The value of b in terms of a0 is equal to 2Z^2πh^2/me^2. To continue, the second derivative of u is needed and equations can be written for the coefficients to solve for E, b, and B/A.
  • #1

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The equation $$\frac{\hbar^2}{2m}\frac{d^2u}{dr^2}-\frac{Ze^2}{r}u=Eu$$ gives the schrodinger equation for the spherically symmetric functions ##u=r\psi## for a hydrogen-like atom.

In this equation, substitute an assumed solution of the form ##u(r)=(Ar+Br^2)e^{-br}## and hence find the values of ##b## and the ratio ##B/A## for which this form of solution satisfies the equation. Verify that it corresponds to the second energy level, with ##E=Z^2/4## times the groud-state energy of hydrogen, and with ##B/A=-Z/2a_0## where ##a_0## is the Bohr radius for hydrogen. What is the value of the coefficient ##b## in terems of ##a_0##?

effort:

given
$$E_1=\frac{mZ^2e^4}{2\hbar^2}$$
and
$$a_0=\frac{4\pi\hbar^2}{Zme^2}\Rightarrow B/A= \frac{-2Z^2\pi\hbar^2}{me^2}$$
$$\frac{\hbar^2}{2m}[(A+2Br)e^{-br}-b(Ar+Br^2)e^{-br}]-\frac{Ze^2}{r}(Ar+Br^2)e^{-br}=\frac{mZ^4e^4}{8\hbar^2}(Ar+Br^2)e^{-br}$$
$$[(\frac{\hbar^2}{2mr}-\frac{b\hbar^2}{2m}-\frac{Ze^2}{r})Ar+(\frac{\hbar^2}{2mr}-\frac{b\hbar^2}{2m}-\frac{Ze^2}{r})Br^2]e^{-br}$$
$$(\frac{\hbar^2}{2mr}-\frac{b\hbar^2}{2m}-\frac{Ze^2}{r})(Ar+Br^2)e^{-br}$$
$$b=\frac{-m^2rZ^2e^4+h^4-2mZe^2h^2}{rh^4}$$
how do you proceed?

The book says the answer is $$b=\frac{Z}{2a_0}=\frac{2Z^2\pi\hbar^2}{me^2}$$

many thanks :bow: edited with progress
 
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  • #2
I forgot to take the second derivative of ##u##!
 
  • #3
The first term on the LHS of the equation should be preceded by a minus sign. Then, as you say, you need the second derivative of u. On the RHS, don't use the expression for E - E is a variable whose value you are to calculate.
Then write equations for the coefficients of the terms in r0, r1 and r2. Solve them to get E, b and B/A. (Note: when comparing with a value of a0 from other sources, remember you are using a units system in which 4πε0 = 1.)
 

1. What are spherically symmetric states in the hydrogen atom?

Spherically symmetric states in the hydrogen atom refer to the electron orbitals in which the probability of finding the electron is the same at any point on a sphere surrounding the nucleus. These states are characterized by the principal quantum number (n) and the angular momentum quantum number (l).

2. How are spherically symmetric states related to the energy levels of the hydrogen atom?

The energy levels of the hydrogen atom are determined by the principal quantum number (n) of the spherically symmetric states. As n increases, the energy level also increases. The angular momentum quantum number (l) determines the sublevels within each energy level.

3. What is the significance of spherically symmetric states in the hydrogen atom?

Spherically symmetric states play a crucial role in understanding the electronic structure of the hydrogen atom. They provide information about the distribution of electrons in the atom and help explain the observed emission and absorption spectra of hydrogen.

4. How do spherically symmetric states relate to the quantum numbers of the hydrogen atom?

The principal quantum number (n) and the angular momentum quantum number (l) correspond to the spherically symmetric states in the hydrogen atom. The magnetic quantum number (m) and the spin quantum number (s) determine the orientation and spin of the electron within each state.

5. Can spherically symmetric states be observed experimentally?

No, spherically symmetric states cannot be observed directly. However, their existence and properties can be inferred from the behavior of electrons in the hydrogen atom, which can be observed experimentally through spectroscopy and other techniques.

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