- #1
Kestrel
- 5
- 0
Hello, I'm writing a program, and while I have a decent understanding of wavefunctions and atomic orbitals, this one appears to be a problem I can't beat.
I'm graphing various properties of atomic orbitals (wavefunction, wavefunction squared, radial probability distribution, and a cross section). I'm trying to find, roughly, the point at which the area under the curve reaches 90% (In otherwords, the integral of the function from 0 to x equals 90% the integral of the function from 0 to infinity). I could, I suppose, have the computer do a Riemann sum and calculate it that way, but I was wondering if there was a simpler way. The math is above me, forgive my ignorance, I'm a senior in high school with a semester of calculus under my belt.
Ideally, there would be a simple factor for each quantum number. For example, if you increase n by 1, the you would just multiply the 90% cutoff by 2, or some such number. Increase l by 1, and you would multiply the 90% cutoff point by 1.4. But I'm sure it's more complicated than that.
Any and all help would be greatly appreciated. If all else fails, I'll do a Riemann sum, but I just don't want to tax the computer any more if there's a simpler way.
Thanks in advance!
Scott
I'm graphing various properties of atomic orbitals (wavefunction, wavefunction squared, radial probability distribution, and a cross section). I'm trying to find, roughly, the point at which the area under the curve reaches 90% (In otherwords, the integral of the function from 0 to x equals 90% the integral of the function from 0 to infinity). I could, I suppose, have the computer do a Riemann sum and calculate it that way, but I was wondering if there was a simpler way. The math is above me, forgive my ignorance, I'm a senior in high school with a semester of calculus under my belt.
Ideally, there would be a simple factor for each quantum number. For example, if you increase n by 1, the you would just multiply the 90% cutoff by 2, or some such number. Increase l by 1, and you would multiply the 90% cutoff point by 1.4. But I'm sure it's more complicated than that.
Any and all help would be greatly appreciated. If all else fails, I'll do a Riemann sum, but I just don't want to tax the computer any more if there's a simpler way.
Thanks in advance!
Scott