The radial wave function of hydrogen atom is R21=124a03(ra0)e−r2a0R_{21}=\frac{1}{\sqrt{24a_0^3}}(\frac{r}{a_0})e^{-\frac{r}{2a_0}}R21=24a031(a0r)e−2a0r
The probability of finding a n = 2, ℓ = 1 electron between a0a_0a0 and 2a02a_02a0 is
P=∫(R21)2r2dr=∫(124a03(ra0)e−r2a0)2r2drP=\intop(R_{21})^2r^2dr =\intop(\frac{1}{\sqrt{24a_0^3}}(\frac{r}{a_0})e^{-\frac{r}{2a_0}})^2r^2drP=∫(R21)2r2dr=∫(24a031(a0r)e−2a0r)2r2dr
=124a05∫r4e−ra0dr=\frac{1}{{24a_0^5}}\int r^4e^{-\frac{r}{a_0}}dr=24a051∫r4e−a0rdr between the limits a0a_0a0 and 2a02a_02a0
On solving the above integration, we get
P =0.04899
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