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Jackson 1.5 Homework Problem Solution Dr. Christopher S. Baird University of Massachusetts Lowell PROBLEM: The time-averaged potential of a neutral hydrogen atom is given by − r = q e 4 0 r 1 r 2 where q is the magnitude of the electronic charge, and α-1 = a0/2, a0 being the Bohr radius. Find the distribution of charge (both continuous and discrete) that will give this potential and interpret your result physically. SOLUTION: The Poisson equation links charge densities and the electric scalar potential that they create. We use it here to find the charge density. We must perform a straight-forward differentiation in spherical coordinates. 2 ∇ =− 0 Expand this in spherical coordinates: 1 ∂ 2∂ 1 ∂2 ∂ sin ∂ 1 r =− 2 2 2 2 2 ∂ r ∂ r ∂ ∂ r r sin r sin ∂ 0 The potential is spherically symmetric, so that the potential depends only on the radial coordinate - the partial derivatives of the potential are all zero, except for the one with respect to the radial component. 1 ∂ 2∂ r =− 2 0 ∂r r ∂r Evaluate the equation explicitly: [ − r 1 ∂ 2 ∂ q e r 2 ∂ r 4 0 r r ∂r [ ] 1 1 r 2 ] =− 0 ] q 1 ∂ 2 ∂ e− r ∂ − r r r 2 e =− 2 0 4 0 r ∂ r ∂r r ∂r 2 [ [ q 1 ∂ 2 1 ∂ − r − r ∂ 1 2 2 − r r e e − r e =− 2 0 4 0 r ∂ r r ∂r ∂r r 2 ] q 1 ∂ 2 2 − r − r 2 − r ∂ 1 − r e r e − r e =− 2 0 4 0 r ∂ r ∂r r 2 q − − r 3 − r q 1 ∂ 2 − r ∂ 1 e e r e =− 2 2 4 0 r 2 4 0 r ∂ r ∂r r 0 3 = −q − r − r q 1 ∂ 2 ∂ 1 e −e r 8 4 r 2 ∂r ∂r r Now we must be careful because 1/r blows up at the origin. Split the last term into two cases: 3 = −q − r − r q 1 ∂ 2 ∂ 1 e −e r 8 4 r 2 ∂r ∂r r = −q − r − r q 1 ∂ 2 ∂ 1 e −e r 8 4 r 2 ∂r ∂r r 3 if r ≈ 0 if r > 0 Away from the origin, 1/r does not blow up and the derivatives can be evaluated normally. The last term ends up equating to zero, so that our equations now becomes: = −q 3 − r − r q 1 ∂ 2 ∂ 1 e −e r 8 4 r 2 ∂r ∂r r = −q 3 − r if r > 0 e 8 if r ≈ 0 At r ≈ 0, we have e− r =1 so that the two cases become: = −q 3 q 1 ∂ 2 ∂ 1 − r 8 4 r2 ∂ r ∂r r = −q 3 − r if r > 0 e 8 if r ≈ 0 Now use the relation: ∇2 1 =−4 r r which when evaluated explicitly becomes: 1 ∂ 2 ∂ 1 r =−4 r ∂r r r2 ∂ r Plug this into the above set of equations: =− q 3 q r if r ≈ 0 8 =− q 3 − r if r > 0 e 8 Because the delta function is zero everywhere except at the origin, and because the first term of the first equation is just the specific r = 0 form of the first term of the second equation, the two cases can be combined into one case: 3 q − r =− e q r for all r 8 This corresponds physically to a positive point charge at the origin with one unit of elementary charge, and a finite cloud of negative charge that decays exponentially, but contains a total charge of one unit of elementary charge. From a time-averaged perspective then, hydrogen in the ground state contains a positive point charge at the center and a circular cloud of negative charge. This is of course only useful for conceptualization purposes, because at atomic sizes the system behaves quantum mechanically, not classically.