Show that the probability density for the ground-statesolution of the one-dimensional Coulomb potentialenergy has its maximum at x = a. Orbital angularmomentumOrbital magneticquantum number|L|= √√(+1)h7.2Angular probabilityP(0, 4) =7.5(1= 0, 1, 2,...)density|0lm,(0)m, ($)|2L₁ = m₁h7.2Orbital magnetic₁ =-(e/2m)L7.6(m,=0,11,12,..., ±1)dipole momentSpatial quantizationL₁₂ m₁cos 0 ==7.2Spin magnetic dipoles = -(e/m)Š7.6-|L|√(+1)momentAngular momentumuncertaintyrelationshipHydrogen quantumnumbersHydrogen energylevelsHydrogen wavefunctionsΔΙΑΦ ΣΑn = 1,2,3,...= 0, 1, 2,...,n-1m₁ =0,11,12,...,±l7.2Spin angular|S|= √√s(s+1)h=7.6momentum√√3/4h (for s = 1/2)Spin magnetic7.3S=mh (m,±2)7.6quantum numberSpectroscopic notationEn=-me 132л²² n²7.37.3Selection rules forphoton emissionNormal Zeeman effects(10), p (1 = 1),d(l=2),f(l= 3),...ΔΙ = +1 Δ = 0,117.77.7, 7.82²R()(0) (0)Radial probabilitydensityP(r) = r²|R(r)|27.4Fine-structure estimateΔλ=AE = mc2a/n= AE = BB7.87.9(a≈1/137)

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Show that the probability density for the ground-state solution of the one-dimensional Coulomb potential energy has its maximum at x = a.