(a) For the following cases, calculate the value of r at which the radial probability density of the hydrogen atom reaches its maximum: (i) n = 1, l = 0, m = 0; (ii) n = 2, l = 1, m = 0; (iii) l = n – 1, m = 0.
(b) Compare the values obtained with the Bohr radius for circular orbits.
(a) Since the radial wave function for n = 1 and l = 0 is R_{10}(r)= 2a^{-3/2}_{0} e^{-r/a_{0} }, the probability density is given by
P_{10}(r)=r^{2} \left|R_{10}(r)\right| ^{2} =\frac{4}{a^{3}_{0}} r^{2}e^{-2r/a_{0} }. (6.259)
(i) The maximum of P_{10}(r) occurs at r_{1}:
\frac{dP_{10}(r)}{dr} \mid _{r=r_{1}} =0\Longrightarrow 2r_{1} -\frac{2r^{2}_{1} }{a_{0} } =0\Longrightarrow r_{1}=a_{0}. (6.260)
(ii) Similarly, since R_{21}(r)=1/(2\sqrt{6}a^{5/2}_{0} )re^{-r/2a_{0} }, we have
P_{21}(r)=r^{2} \left|R_{21}(r)\right| ^{2} =\frac{1}{24a^{5}_ {0} } r^{4} e^{-r/a_{0} }. (6.261)
The maximum of the probability density is given by
\frac{dP_{21}(r)}{dr} \mid _{r=r_{2}} =0\Longrightarrow 4r^{3}_{2} -\frac{r^{4}_{2}}{a_{0}} =0\Longrightarrow r_{2}=4 a_{0}. (6.262)
(iii) The radial function for l = n – 1 can be obtained from (6.170):
R_{nl}(r)=-\left(\frac{2}{na_{0}} \right) ^{3/2} \sqrt{\frac{(n-l-1)!}{2n[(n+l)!]^{3} } } \left(\frac{2r}{na_{0}} \right) ^{l} e^{-r/ na_{0} }L^{2l+1}_{n+l} \left(\frac{2r}{na_{0}} \right) (6.170)
R_{n(n-1)}(r)=-\left(\frac{2}{na_{0}} \right) ^{3/2}\frac{1}{\sqrt{2n[(2n-1)!]^{3} } } \left(\frac{2r}{na_{0}} \right) ^{(n-1)} e^{-r/na_{0} }L^{2n-1}_{2n-1}\left(\frac{2r}{na_{0}} \right). (6.263)
From (6.159)
L^{N}_{k} (r)=\frac{d^{N} }{dr^{N} } L_{k} (r), (6.159)
and (6.160)
L_{k} (r)=e^{r} \frac{d^{k} }{dr^{k} } (r^{k}e^{-r} ). (6.160)
we can verify that the associated Laguerre polynomial L^{2n-1}_{2n-1} is a con-stant, L^{2n-1}_{2n-1}(y)=-(2n-1)!. We can thus write R_{n(n-1)}(r) as R_{n(n-1)}(r)=A_{n} r^{n-1} e^{-r/na_{0}}, where A_{n} is a constant. Hence the probability density is given by
P_{n(n-1)} (r)=r^{2} \left|R_{n(n-1)}(r)\right| ^{2} =A^{2} _{n} r^{2n} e^{-2r/na_{0}}. (6.264)
The maximum of the probability density is given by
\frac{dP_{n(n-1)} (r)}{dr} \mid _{r=r_{n} } =0\Longrightarrow 2nr^{2n-1}_{n} -\frac{2r^{2n}_{n} }{na_{0}} =0\Longrightarrow r_{n}=n^{2} a_{0}. (6.265)
(b) The values of r_{n} displayed in (6.260), (6.262), and (6.265) are nothing but the Bohr radii for circular orbits, r_{n}= n^{2} a_{0}. The Bohr radius r_{n}= n^{2} a_{0} gives the position of maximum probability density for an electron in a hydrogen atom.
