For a hydrogen-like atom, the radial part of a wavefunction is proportional to sigma² * e^(-sigma/2) and the angular part is proportional to sin(theta)*cos(theta)*cos(phi), where sigma = 2r/a0. Using a hypothetical assignment where px, py, pz have m = +1, -1, 0, and dxy, dyz, dzx, d(x²-y²)

Question

For a hydrogen-like atom, the radial part of a wavefunction is proportional to sigma² * e^(-sigma/2) and the angular part is proportional to sin(theta)*cos(theta)*cos(phi), where sigma = 2r/a0. Using a hypothetical assignment where px, py, pz have m = +1, -1, 0, and dxy, dyz, dzx, d(x²-y²), dz² have m = +2, -2, +1, -1, 0 respectively, find the value of (n + l + m) for this orbital.

Accepted Answer

6. The radial function R ~ sigma² * e^(-sigma/2). For hydrogen orbitals, R_nl ~ r^l * polynomial * e^(-r/(n*a0)). With sigma = 2r/a0, sigma = 2r/a0 so r/a0 = sigma/2. e^(-sigma/2) = e^(-r/a0). For n=3 level: e^(-r/(3a0))... wait, for 3d: R₃₂ ~ r² * e^(-r/(3a0)). In terms of sigma=2r/a0: r = sigma*a0/2, e^(-r/(3a0)) = e^(-sigma/6). But the given function has e^(-sigma/2). For n=3: sigma = 2Zr/(n*a0) = 2r/a0 only if n=1... Actually in general sigma = 2Zr/(n*a0). For Z=1: sigmaₙ = 2r/(n*a0). The given sigma = 2r/a0 means n=1 in standard notation. BUT if we use sigma = 2r/a0 universally (not dividing by n), then for 3d orbital: R ~ (r/a0)² * e^(-r/(3a0)) = (sigma/2)² * e^(-sigma/6). That doesn't

Solution

Related JEE Advanced Physics questions