Phy5645/HydrogenAtomProblem3: Difference between revisions

Latest revision as of 13:43, 18 January 2014

(a) The ground state wave function for hydrogen is

$\psi ={\frac {e^{-r/a}}{\sqrt {\pi a^{3}}}},$

so that

$\langle r^{n}\rangle ={\frac {1}{\pi a^{3}}}\int \ r^{n}e^{-2r/a}r^{2}\sin {\theta }\,dr\,d\theta \,d\phi ={\frac {4}{a^{3}}}\int _{0}^{\infty }\!r^{n+2}e^{-2r/a}\,dr={\frac {4}{a^{3}}}\left({\frac {a}{2}}\right)^{n+3}\int _{0}^{\infty }\!x^{n+2}e^{-x}\,dx={\frac {(n+2)!}{2^{n+1}}}a^{n}.$

In particular, $\langle r\rangle ={\tfrac {3}{2}}a$ and $\langle r^{2}\rangle =3a^{2}.$

(b)

The probability of finding the electron within a spherical shell of thickness $dr\!$ is

$P=|\psi |^{2}\cdot 4\pi r^{2}\,dr={\frac {4}{a^{3}}}r^{2}e^{-2r/a}\,dr=p(r)\,dr.$

The probability per unit length $p(r)$ is then

$p(r)={\frac {4}{a^{3}}}r^{2}e^{-2r/a}.$

The most probable value of $r\!$ is then found by maximizing $p(r):\!$

$0={\frac {dp}{dr}}={\frac {4}{a^{3}}}\left[2re^{-2r/a}-r^{2}\left({\frac {2}{a}}e^{-2r/a}\right)\right]={\frac {8r}{a^{3}}}e^{-2r/a}\left(1-{\frac {r}{a}}\right)$

The only non-trivial solution is $r=a.\!$ We know that the probability per unit length goes to zero at $r=0\!$ and as $r\to \infty ,\!$ so that $r=a\!$ must be a maximum.

Back to Hydrogen Atom

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-(Submitted by team 5. This is based on problems 4.13 and 4.14 from Quantum Mechanics by Griffiths)
+'''(a)''' The ground state wave function for hydrogen is
-'''Solution'''
+<math>\psi = \frac{e^{-r/a}}{\sqrt{\pi a^3}},</math>
-<math>\psi = \frac{e^{\frac{-r}{a}}}{\sqrt{\pi a^3}}</math>
+so that
-so <math> <r^n> = \frac{1}{\pi a^3} \int\ r^n e^{\frac{-2r}{a}} r^2 sin{\theta}\, dr\,d\theta\,d\phi = \frac{4\pi}{\pi a^3} \int_0^\infin \! r^{n+2} e^{\frac{-2r}{a}} \, dr</math>
+<math>\langle r^n\rangle = \frac{1}{\pi a^3} \int\ r^n e^{-2r/a} r^2\sin{\theta}\, dr\,d\theta\,d\phi = \frac{4}{a^3} \int_0^\infin \! r^{n+2} e^{-2r/a} \, dr=\frac{4}{a^3}\left (\frac{a}{2}\right )^{n+3}\int_0^\infin \! x^{n+2} e^{-x} \, dx=\frac{(n+2)!}{2^{n+1}}a^n.</math>
-<math> <r> = \frac{4}{a^3} \int_0^\infin \! r^{3} e^{\frac{-2r}{a}} \, dr = \frac{4}{a^3} 3! (\frac{a}{2})^4 = \frac{3}{2} a </math>
+In particular, <math>\langle r\rangle=\tfrac{3}{2}a</math> and <math>\langle r^2\rangle=3a^2.</math>
-<math> <r^2> = \frac{4}{a^3} \int_0^\infin \! r^{4} e^{\frac{-2r}{a}} \, dr = \frac{4}{a^3} 4! (\frac{a}{2})^5 = 3 a^2 </math>
+'''(b)'''
-'''(B)''' What is the most probable value of r, in the ground state of hydrogen?
+The probability of finding the electron within a spherical shell of thickness <math>dr\!</math> is
-'''Solution'''
+<math> P = |\psi|^2\cdot4\pi r^2\,dr = \frac{4}{a^3}r^2e^{-2r/a}\,dr = p(r)\,dr.</math>
-<math> \psi = \frac{1}{\sqrt{\pi a^3}} e^{\frac{-r}{a}} </math>
+The probability per unit length <math>p(r)</math> is then
-<math> P = \mid \psi \mid^2 4 \pi r^2 dr = \frac{4}{a^3} e^{\frac{-2r}{a}} r^2 dr = p(r) dr; p(r) = \frac{4}{a^3} r^2 e^{\frac{-2r}{a}} </math>
+<math>p(r) = \frac{4}{a^3} r^2 e^{-2r/a}.</math>
-<math> \frac{dp}{dr} = \frac{4}{a^3} [2r e^{\frac{-2r}{a}} + r^2 ( \frac{-2}{a} e^{\frac{-2r}{a}})] = \frac{8r}{a^3} e^{\frac{-2r}{a}} (1 - \frac{r}{a}) = 0 \implies r = a </math>
+The most probable value of <math>r\!</math> is then found by maximizing <math>p(r):\!</math>
-Back to [[Hydrogen Atom]]
+<math>0=\frac{dp}{dr} = \frac{4}{a^3}\left [2r e^{-2r/a}-r^2\left ( \frac{2}{a} e^{-2r/a}\right )\right ] = \frac{8r}{a^3} e^{-2r/a}\left (1 - \frac{r}{a}\right )</math>
+The only non-trivial solution is <math> r = a.\!</math>  We know that the probability per unit length goes to zero at <math>r=0\!</math> and as <math>r\to\infty,\!</math> so that <math>r=a\!</math> must be a maximum.
+Back to [[Hydrogen Atom#Problems|Hydrogen Atom]]

Phy5645/HydrogenAtomProblem3: Difference between revisions

Latest revision as of 13:43, 18 January 2014

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