Answer to Question #85013 in Quantum Mechanics for akash singh

Question #85013
Obtain the most probable value and expectation value of r for the ground state of a
hydrogen atom.
1
Expert's answer
2019-03-11T11:04:08-0400

The wave function of an electron in the ground state of hydrogen atom


ψ=1πa03er/a0\psi=\frac{1}{\sqrt{\pi a_0^3}}e^{-r/a_0}

The probability density


ρ=4πr2ψ2=4r2a03e2r/a0\rho=4\pi r^2|\psi|^2=\frac{4r^2}{a_0^3}e^{-2r/a_0}

dρdr=8ra03[1ra0]e2r/a0=0\frac{d\rho}{dr}=\frac{8r}{a_0^3}\left[1-\frac{r}{a_0}\right]e^{-2r/a_0}=0

So, the most probable value value of r 


rmp=a0r_{\rm{mp}}=a_0

The expectation value of r

<r^>=0rρdr=04r3a03e2r/a0=32a0<{\hat r}>=\int_0^{\infty} r \rho dr=\int_0^{\infty}\frac{4r^3}{a_0^3}e^{-2r/a_0}=\frac{3}{2}a_0


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