(a) Calculate the expectation value 〈r〉_{21} for the hydrogen atom and compare it with the value r at which the radial probability density reaches its maximum for the state n = 2, l = 1.
(b) Calculate the width of the probability density distribution for r.
(a) Since R_{21} (r)=re^{-r/2a_{0} } /\sqrt{24a^{5}_{0} } the average value of r in the state R_{21} (r) is
〈r〉_{21}=\frac{1}{24a^{5}_{0}} ∫^{\infty }_{0} r^{5} e^{-r/a_{0} }dr=\frac{a_{0}}{24} ∫^{\infty }_{0}u^{5} e^{-u} du= \frac {120a_{0}}{24} =5a_{0}; (6.266)
in deriving this relation we have made use of ∫^{\infty }_{0} x^{n} e^{-x} dx =n!.
The value r at which the radial probability density reaches its maximum for the state n = 2, l = 1 is given by r_{2} =4a_{0}, as shown in (6.262).
\frac{dP_{21}(r) }{dr} \mid _{r=r_{2}} =0\Longrightarrow 4r^{3}_{2} -\frac{r^{4}_{2} }{a_{0}} =0\Longrightarrow r_{2}= 4a_{0}. (6.262)
What makes the results r_{2} =4a_{0} and 〈r〉_{21} =5a_{0} different? The reason that 〈r〉_{21} is different from r_{2} can be attributed to the fact that the probability density P_{21}(r) is asymmetric about its maximum, as shown in Figure 6.5. Although the most likely location of the electron is at r_{0}=4a_{0}, the average value of the measurement of its location is 〈r〉_{21} =5a_{0}.
(b) The width of the probability distribution is given by \Delta r=\sqrt{〈r^{2} 〉_{21}-〈r〉^{2}_{21} } , where the expectation value of r^{2} is
〈r^{2} 〉_{21}=∫^{\infty }_{0}r^{4} R^{2}_{21} (r)dr =\frac{1}{24a^{5}_{0} } ∫^{\infty }_{0}r^{6}\exp \left(-\frac{1}{a_{0}}r \right) dr=\frac{6!a^{7}_{0}}{24a^{5}_{0}} =30a^{2}_{0}. (6.267)
Thus, the width of the probability distribution shown in Figure 6.5 is given by
\Delta r_{21}=\sqrt{〈r^{2} 〉_{0}-〈r〉^{2}_{0}} =\sqrt {30a^{2}_{0}-(5a_{0} )^{2} } =\sqrt{5} a_{0}. (6.268)
