04 EM – 051 – pr 05: The time-averaged potential of a neutral hydrogen atom is given by:

( )12

0

exp( )( ) 1

4

q rr r

r

αα

πε

−Φ = + [I.1]

…where “q” is the magnitude of the electronic charge, and 1 1 1

02 2 [ ]a Bohr radiusα − = = .

Find the distribution of charge (both continuous and discrete) that will given this potential. Interpret your result

physically.

Gauss’s law in integral form and divergence theorem results in differential Gauss-law:

0/ρ ε∇• =E� �

[I.2]

That the electric field is conservative/irrotational implies = −∇ΦE��

, and this makes [I.2] into:

2 2 2

0 0 , ,

0

( ) r θ φ

ρρ ε ε

ε∇ • = −∇ • ∇Φ = −∇ Φ = → = − ∇ Φ = − ∇ ΦE� � � � � ��

[I.3]

Looking at the time-averaged potential [I.1], we see it is a function only of “r”, so angular derivatives of the full Laplacian

(as in [I.3]) vanish. Thus, we only need the radial component of Laplacian:

2 2

2

1r

rr r r

∂ ∂Φ ∇ Φ =

∂ ∂

�

[I.4]

Rearrange the given potential [I.1] so that the trivial 04

q

πε is away from the functions of “r”,

( ) ( )0 1 1 1 12 2

4( ) 1 ( ) ( )

r r

r rr e r e f r g r

q

α απεα α− −Φ = + = + = ⋅ [I.5]

Thus, we are to apply 2

r∇�

of [I.4] to the product of functions ; what results are binomial coefficients in ∇�

,

( ) ( ) ( )2 2 2 2 2( ) 2f g f g f g f g f g f g f g g f f g f g f g∇ ⋅ = ∇ • ∇ ⋅ = ∇ • ∇ ⋅ + ⋅∇ = ∇ ⋅ + ⋅∇ + ∇ • ∇ + ∇ • = ∇ ⋅ + ⋅∇ + ∇ • ∇

� � � � � � � � � � � � � � �

[I.6]

Calculating f∇�

, g∇�

, 2

f∇�

, and 2g∇�

in which, by [I.5], 1 12r

f α= + and rg e α−= , and using [I.4] to handle “g” and

2 31 4 ( )r rπ δ∇ ≡ − ⋅ to handle “f”,

( )2

1 1 1ˆ ˆ ˆ ˆ

2

r rf g e er r r r

α αα α− −∂ − ∂ ∇ = + = ∇ = = −

∂ ∂ r r r r

� �

[I.7]

2 2 31 1

4 ( ) 4 ( )2

f rr

α π δ π δ

′∇ = ∇ + = − ⋅ − = − ⋅

x x� �

[I.8]

( ) ( ) ( )2 2 2 2 2 22 2 2

12r r r r r

rg r e r e re r e er r r r r r

α α α α ααα αα α− − − − −∂ ∂ − ∂ −

∇ = = = − = − ∂ ∂ ∂

�

[I.9]

Putting [I.7], [I.8], and [I.9] together a la [I.6] to constitute the Laplacian of the product [I.5],

( ) ( ) ( ) ( )

( ) ( )

2

2

2 2

2 2 30 21 1 12

2 3 3 3 31 1 12 2

4ˆ ˆ( ) 4 ( ) 2

2 4 ( ) 2 4 ( )

r r r

r r r

r r

r rr r

r e e r eq

r e r e

α α αα

α αα α α

πεα α π δ α

α α π δ α π δ

− − −−

− −

∇ Φ = + − − ⋅ + ⋅ • −

= − + − − ⋅ + = − ⋅

r r [I.10]

The charge density by [I.3] is,

( ) ( )2 3 3 3 31 10 0 2 2

0

4 ( ) 4 ( )4 4

r rq qr e r eα αρ ε ε α π δ α π δ

πε π− −−

= − ∇ Φ = − − ⋅ = − ⋅�

[I.11]

Realizing 0 0

0 0( ) ( )

x xax

x xx e dx x dx

δ δ

δ δδ δ

+ +−

− −⋅ ⋅ = ⋅∫ ∫ , [I.11] simplifies to,

( )3

3 3 312

4 ( ) ( )4 8

r rqe r q r eα αα

ρ α π δ δπ π

− − −= − ⋅ = −

[I.12]

Clearly: 3( )q rδ⋅ is the charge density of the kernel-nucleus, and

3

8

rq e

ααπ

−− is the charge density of the surrounding

electron cloud.

checking our answer: The hydrogen atom is charge neutral, so [I.12] should integrate to zero over all space; using 3 2

4d x r drπ′ = ⋅ ,

( ) ( )3 3 3 3 3 3 3 21 18 2

0 0( ) ( ) ( )r rQ d x q e d x q d x e r drα α

πρ δ α δ α∞ ∞

− −′ ′= ⋅ = − = ⋅ − ⋅∫ ∫ ∫ ∫x x x [I.13]

To integrate 2

0

re r drα∞

− ⋅∫ , our instructor Dr. Chabysheva invites us to remember the definition of the gamma function,

1

0( ) ( 1)!n xn x e dx n

∞− −Γ ≡ ⋅ = −∫ [I.14]

Thus, after effecting 3 3( ) 1d xδ ⋅ =∫ x , [I.13] becomes,

( ) ( ) ( ) ( )3

3 3 11 1 1 12 2 2

0(1) ( ) 1 (3) 1 2! 1 1 0rQ q e r d r q q qα

αα α

∞− −= − ⋅ = − Γ = − = − =∫ [I.15]

Plotting this charge density [I.12] and letting 2200

( )r

r eδ −≈ ,

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.91

0.5

0

0.5

1

exp x−( )−

exp 200− x2

⋅( )

exp 200− x2

⋅( ) exp x−( )−

x

- charge density units: 4*pi*q/epsilon[0]

- length units: Bohr radii, "alpha"

proton alone

proton + electron cloud

electron cloud alone

[I.16]

Alternate solution

The shorter way, which is in our best interest for the exam, would be: first consider applying 2

0ε− ∇ to [I.5] with r > 0,

( )( )( )( )

( ) ( ) ( )( )

( )

2

1 12 2 22 1 1 1

0 0 0 22 2

2 2 2 21 10 02 22 2

2 2 2 31 10 0 0 02 22

( )1 1

1 11 1 1

1 ( )4

r

r rr

r r

r r

r r

er r e e

r r r r r

e r r e r r rr r r

re r e

r q

αα α

α α

α α

αε ε ε α α

ε α α ε α α α α α

ρε αε α α ε πε

−− − −

− −

− −

∂ +∂ ∂− ∇ Φ = − = − − + +

∂ ∂ ∂

∂= + + = − + + + +

∂

− ∇ Φ = − = − =

[I.17]

Solving for ( )rρ in [I.17], we get,

31

8( )r

r q eα

πρ α −= − [I.18]

So, blind application of the operator 2

0ε− ∇ yields the second term in [I.12] (in units where0

14

1πε = ). Now, consider the

limit 0r → , where we may let 1x

e x≈ + . This results in,

2 2 2 2 2 20 1 1 1 1 1 1 1 1

0 0 0 02 2 2 2 2

4(1 )( ) ( ) ( )

r r rr r r

q

πεε ε α α ε α α α ε α α

Φ− ∇ = − ∇ − + = − ∇ + − − = − ∇ − − [I.19]

In [I.19], for 0r → , we have 1r

r>> , so we may discard that term too. From there, we effect the same rigamarole at the

beginning of [I.17], then, except we bear in mind the formula 2 1( ) 4 ( )r rπ δ∇ ≡ − ⋅ , and [I.19] becomes,

( )2 2 2 2010 0 0 0 022 2

1( 0 1) 4 ( ) 0 4 ( )

2r r r r r

r r r r r

α αεε ε ε α πε δ πε δ

∂ ∂ ∂ − ∇ Φ = − ∇ + ⋅ − = ⋅ + ⋅ = ⋅

∂ ∂ ∂ [I.20]

Bam: [I.18] and [I.20] constitute both charge densities!