The normalized wave function for a hydrogen atom in the 1s state is given by psi(r)=1/(sqrt(sqrt(pi) alpha0))*e^(-r/(alpha0)) where a0 is the Bohr radius, which is equal to 5.29 x 10^11 m. What is the probability of finding the electron at a distance greater than 3.9 a0 from the proton?

(Note: This problem has been edited from its original version, to correct an obvious error in the units of measurement and the normalization of the wavefunction that made it unsolvable.)We are given this wavefunction:psi(r) = 1/sqrt(alpha_0 sqrt(pi)) e^{-r/alpha_0}
The probability density of finding the atom at a given point is given by the absolute square of the wavefunction at that point:
p(r) = abs(psi(r))^2 = 1/(sqrt(pi) alpha_0) e^{-r^2/alpha_0^2}
Normally, to get the overall probability of it being at or further than a particular point x, we would integrate that probability density from x to infinity: P(x gt= r) = int_{x}^{infty} p(r) dr = int_{x}^{infty} 1/(sqrt(pi) alpha_0) e^{-r^2/alpha_0^2} But in this case, that integral is not possible to solve exactly. Fortunately, there's a trick we can use. We can recognize that it is in fact a Gaussian function (also called a normal distribution or bell curve). N(mu, sigma) = 1/(sigma sqrt(2 pi)) e^{-(x-mu)^2/(2 sigma^2)} mu = 0 is fairly easy to see. But what about sigma ?We want sigma sqrt(2pi) = sqrt(pi) alpha_0 , so sigma = alpha/sqrt(2) That had better work in the other part of the equation, and indeed it does:1/(2 sigma^2) = 1/alpha_0^2 From here, we can just look up the result in a table, because Gaussian functions have well-documented numerical solutions written in statistical tables. To be 3.9 a0 from the origin, we need to be 5.5 standard deviations from zero. The probability of that can be read off the table, as 2*10^-8.
http://mathworld.wolfram.com/NormalDistribution.html