Question

show all work

expression for determining the average distance between the electron and the nucleus. ($ pts) an electron in the 3d orbital of a Sc atom 56. Calculate the distance from the nucleus at which the electron is most likely to be found. You can leave your answer in terms of ay and Z(13 pts) Se. Set up (but do not evaluate) the integral for determining the probability that this electron wil be found at a distance between 25 pm and 50 pm from the nucleus. Provide enough information to be able to do the integration directly (Gie. include limits of integration, and do not leave your answer in terms of generic y and dt). (4 pts) The avernge dlichance number of e-.ae is in the _

Answer 1

ga,The expression or -determining-the average dletexm distance betoeen 寸he electrna and nucleus. For an elecron in the axound stato ot an actom tte electrostatte attratws can be wnitea Thu fore must be to the certn ptal 2 TT 4-trs 2- an 2 1 2 Dist

prom the nucleus at which the distamce electon is mo st likel t be tound is ven bu the expe ctatuwn value o 군 for duffevent nand vabies alifferent

<x7 = C).as/ 3,-. 242tI)) = 9Aa / 3-1) 2-I 2(3) 7-12 3

Distance between 25pm and 5D 3 ca SD pm -2군 3 ap 25 pm