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?

Expert Answers

An illustration of the letter 'A' in a speech bubbles

(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...

(The entire section contains 260 words.)

Unlock This Answer Now

Start your 48-hour free trial to unlock this answer and thousands more. Enjoy eNotes ad-free and cancel anytime.

Start your 48-Hour Free Trial
Approved by eNotes Editorial Team