Spherically symmetric states in the hydrogen atom

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Homework Statement:
psb
Relevant Equations:
psb
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
 
Last edited:

Answers and Replies

  • #2
docnet
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I forgot to take the second derivative of ##u##!
 
  • #3
mjc123
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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.)
 

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