The probability P that an electron is a distance r from the nucleus is called the radial distribution function. The hydrogen atom’s is electron radial distribution function is

Where a_{0} is a constant called the Bohr radius (a_{0} = 0.5292 A or a_{0} = 0.5292 x 10^{-10} m). Below Figure plots the hydrogen atom’s is electron radial distribution function with a_{0} = 0.5292 A.

What is the probability of finding the electron at a radius of less than 2a_{0}, i.e. integrate the above equation from r = 0 to r = 2a_{0}

Make use of the standard integral below.

**ʃ x ^{2} e^{bx} dx = e^{bx} [ x^{2}/b – 2x/b^{2} – 2/b^{3}] + C**

So,

The required probability comes from integrating the radial distribution function between r = 0 and r = 2a_{0}. Firstly, take the constant term (4/a_{0}^{3}) outside the integral.

The integral is now recognizable in terms of the standard integral below.

**ʃ x ^{2} e^{bx} dx = e^{bx} [ x^{2}/b – 2x/b^{2} – 2/b^{3}] + C**

Using the substitutions x = r and b= —2/a_{0} and not forgetting the constant (4/a_{0}^{3}), the probability integral is

The best way to deal with this complicated looking integral is to clean-up the denominators.

We now substitute in the two limits for our definite integral and the integration constant cancels out.

As e^{0} = 1, and cancelling out terms we obtain,

So there is 0.7619 fractional probability *(i.e. *fraction of unity) that the electron will be within 2a_{0} of the nucleus for the is atomic orbital of the hydrogen atom. This is the same as saying there is a 76.19% chance that the 1s electron is between the nucleus and a distance of 2a_{0}. See Figure for the radial distribution function and this Figure for the 3-dimensional is atomic orbital drawn to enclose 90% of the total probability.