Modern Physics. Unit 6: Hydrogen Atom - Radiation Lecture 6.1: The Radial Probability Density. Ron Reifenberger Professor of Physics Purdue University

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1 Mdern Physics Unit 6: Hydrgen Atm - Rditin Lecture 6.1: The Rdil Prbbility Density Rn Reifenberger Prfessr f Physics Purdue University 1

2 Prbbility Density Prbbility Density * ΨΨ = Ψ In 1-D, the prbbility f finding prticle between x nd x+dx is Ψ In -D, the prbbility f finding prticle between (r,θ,φ) nd (r+dr,θ+dθ,φ+dφ) is Ψ * ( x) Ψ ( ) * (, θφ, ) Ψ (, θφ, ) x dx r r dv Prbbility must be pure number - implictins fr units f Ψ

3 The vlume element dv in sphericl crdintes The prlellpiped pprches rectngulr bx s dr, dθ, dφ, s the vlume element is just The prbbility P(r, θ, φ ) tht yu find n electrn t sme pint specified by r=r, θ=θ, φ=φ in vlume element dv=r sinθ drdθdφ is Ψ dv where Ψ=R n,l (r ) Θ l,m (θ ) Φ m (φ ).

4 Defining the Rdil Prbbility Density The prbbility tht yu find the electrn in shell between rdius r nd r+ r (r is the distnce frm the nucleus) is given by integrting ver ll ngles: π π ( ) = n, Θ, m( ) sin Φm( ) P r dr r R dr θ θdθ ϕ dϕ If the vrius wvefunctins hve been prperly nrmlized t unity, then () = n, 1 1 P r dr r R dr The rdil prbbility density = prbbility f finding n electrn in shell f thickness r t rdius r is given simply by P() r dr = r R dr

5 Exmple I: Clculte the rdil prbbility density fr the quntum stte Ψ n,l,m when n=; l=1, m= Using the definitin f the rdil prbbility density P() r dr = r Rn, dr R,1 1 1 = r r 1 1 r r R,1 = r e e 4 r 1 r r 1 r e 4 = = 8 r e r

6 Wrking it ut in detil gives the sme nswer π π ( ) = n, Θ, m( ) sin Φm( ) P r dr r R dr θ θdθ ϕ dϕ R r r n, e, m θ m π = Θ = cs Φ = r π r 1 4 π P( r) dr = r e dr cs θsinθdθ π 4 π 4 r r θ r 4 e dr 4 1 cs 1 r = = cs π cs e dr π r r ( 1) (1) 4 e dr 1 = r r r r = e dr = e dr r r r 1 = 4 e d 4 in dimensinless frm must integrte t unity if wvefunctins re prperly nrmlized π

7 Exmple II: Plt the rdil prbbility density fr the quntum stte Ψ n,l,m when n=1; l=, m= R r = e r / 1, () R 1, ( r ).5 1,,.6 [( ) 1/ ]*R1, ( r ) R r r R () r 1, 1, () (r/ ) Prbbility Density.4.. (r\ ) * R 1,( r ) r/

8 Exmple III: Plt the rdil prbbility density fr the quntum stte Ψ n,l,m when n=; l=, m= 1 r R,() r = 1 e R, ( r ) r/.8,,.5 [( ) 1/ ]*R, ( r ) Prbbility Density R () r, r R () r, (r\ ) * R,( r ) r/ Phse f wvefunctin (+,-): imprtnt fr chemicl bnding Rdil nde

9 r R () r, Dt density representtin f rdil prbbility density r R, ( r ) [( ) 1/ ]*R, ( r ) ,, Prbbility Density R () r, (r\ ) * R,( r ) r/

10 Exmple IV: If n electrn is in the n=, l=1 stte, wht is the mst likely distnce frm the rigin tht yu will find the electrn? 1 r R,1() r = e 4 r/ Prbbility f finding electrn t r: 1 r Pr () = r R,1() r = r e 4 r /. The mst likely distnce is where the prbbility density hs mximum..5 Find r where P(r) is mximum: dp() r 1 d 4 r / ( re = ) = dr 4 dr = re r e 5 + = 4 r / 4 r / 4 [( ) 1/ ]*R,1 ( r ) R,1 ( r ) Prbbility Density.1 r / 4 e r = 4r r = = 4..,1, (r\ ) *R,1( r ) r/ 1

11 Up Next: Angulr Mmentum

12 Exmple V: An electrn is in the n=1, l= stte. Wht is the prbbility f finding the electrn t distnce r < frm the nucleus? n=1, l= stte: R r = e r / 1,() Prbbility f finding electrn between r nd r+dr: r / P() r dr = r R1,() r dr = r e dr Integrte t find prbbility between nd : 4 r / P( r =, r = ) = P( r) dr = r e dr r 1 x let x P x e dx = = ( x ) ( ( )[ ] [ ]) ( ) P=.5 e x + x+ =.5 e 1 ( 1) = =.5 4 1

13 Exmple VI: An electrn is in the n=, l=1, m= stte. Wht is expecttin vlue fr r,1? Ψ ( r, θφ, ) = R ( ry ) ( θφ, ) n,, m n,, m Ψ ( r, θφ, ) = R ( ry ) ( θφ, ),1,,1 1, < r >= R () ry (, θφ) r R () ry (, θφ) dv * *,1,1 1,,1 1, = r r R,1() r dr 1 r r 1 r r = r e dr ( ) 4 = 4 let x = r dx = dr 1 = 4 ( ) 1 5 x x 5 4 = x e dx e 4 = = 1 = x x e dx ( x 5x x 6x 1x 1) 1 r R,1() r = e 4 r/ e r dr 1

14 Exmple VII: An electrn is in the n=, l=1, m= stte. Wht is the expecttin vlue fr 1/r? Ψ ( r, θφ, ) = R ( ry ) ( θφ, ) n,, m n,, m Ψ ( r, θφ, ) = R ( ry ) ( θφ, ) 1 r let,1,,1 1, = = 1 r 1 * * R,1() ry1,(, θφ) R,1() ry1,(, θφ) dv r r R,1() r dr r r r r r = r e dr = r e dr 4 4 x = r dx = dr 1 = 4 ( ) x x e dx 1 r R,1() r = e 4 r/ 1 1 x x 1 x e dx e ( x x 6x 6) = = = (s expected frm Ex. IV) 14

15 Exmple VIII: An electrn is in the n=, l=1, m= stte. Wht re the expecttin vlues fr U(r) nd K? U(r) e -, m e n=; E=-.4 ev e +, m p r= 15

16 <U(r)> is esy: Ur () Cnservtin f energy gives <K>: 1 e e 1 e 1 = = = 4πε r 4πε r 4πε 4 1 e 1 e 1 4πε n e ; 4πε n n= 4πε 4 mee K = E U = m = 1 e me 1 e 1 me 4πε me 4 4πε 4 = + em e e 4πε m e 1 1 = me + 8 m = + = 8m 4m 8m e e e (see Ex. VII) U depends n e; K depends n m 16

17 Exmple IX: Des the previus result gree with Viril Therem (see Lecture 1.)? if V(r) 1/r, then <K> time ver. =-½<V> time ver. 1 Des K = V? 1 1 e 1 = 8me 4πε 4 1 m e = 4πε e = 4πε m ee Yes!

18 Exmple X: An electrn is in the n=1, l=, m= stte. Shw tht the prbbility f finding the electrn between r= nd r= is unity. Ψ ( r, θφ, ) = R ( ry ) ( θφ, ) n,, m n,, m Ψ (, r θφ, ) = R () r Y (, θφ) = R () r Θ 1,, 1,, 1, e 1 1 r / = π π π (,, ) n, ( ), m( ) sin θ, m( φ) π π r / 1 1 r e dr sinθ dθ π ( θ ) Φ,, P = P r θφdv = r R r dr Θ θ θd Φ dφ 4 = ( φ) π π r / r / r e dr [ csθ ] [ φ] r e π = = 4 r / r 4 x 1 4 dφ P = r e dr let x = dr = dx x ( ) P = x e dx = e x + x + = 1 dr 18

19 Nte: t sve time, use n-line integrtr; it s free. See: