Figure 6.12b shows the probability of finding a hydrogen 1s electron at various distances from the nucleus. To create the graph in this figure, the electron cloud is first divided into a series of thin concentric shells about the nucleus and then the probability of finding the electron in each shell is evaluated. The volume of each shell is given by the equation V = 4πr²(d), where d is the thickness of the shell and r is the distance of the shell from the nucleus. The probability of finding the electron in each shell is Probability = 4πr2ψ²(d) where ψ is the 1s wavefunction for hydrogen (a_o in this equation is 52.9 pm).

(a) The most probable distance for a 1s electron in the hydrogen atom is at 52.9 pm. Evaluate the probability of finding the electron in a concentric shell 1.0 pm thick at this distance from the nucleus.

(b) Calculate the probability of finding the electron in a shell 1.0 pm thick at distances from the nucleus of 0.50 a_o and 4 a_o. Compare the results with the probabilities at a_o. Are these probabilities in accord with the surface density plot shown in Figure 6.12b?

Data given in Figure 6.12

Transcribed Image Text:

= 1 ()1/2 e-r/ao