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Griffiths 4.14: What is the most probably value of r, in the ground state of hydrogen? (Hint: First you must figure out the probability that the electron would be found between r and r+dr.

Here is the posted solution:

http://www.glue.umd.edu/~syang9/problem%204.14%20solution.PNG [Broken]

I don't understand the approach; why does multiplying the probability density by [tex] \[

4\pi r^2 dr\] [/tex] give the probability that the electron will be found between r and r+dr? Why do we differentiate p(r)? In general, I thought the way to find the probability of a particle being in a particular energy eigenstate was to take the inner product of the general wavefunction with the energy eigenstate..

[tex]

\[

\left| {c_n } \right|^2 = \int_{ - \infty }^\infty {\Psi ^* (\overrightarrow {\bf{r}},t) \cdot \psi (r){\rm{ }}dr}

\]

[/tex]

But the general wave function is a sum which involves [tex] c_n [/tex].. so I don't understand what governs which approach to take..

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# Hydrogen probably homework

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