1 answer

Consider an electron within the ls orbital of a hydrogen atom. The normalized probability of finding...

Question:

Consider an electron within the ls orbital of a hydrogen atom. The normalized probability of finding the electron within a sp

Consider an electron within the ls orbital of a hydrogen atom. The normalized probability of finding the electron within a sphere of a radius R centered at the nucleus is given by normalized probability = [az-e * (až + 2a, R+ 2R)] where a, is the Bohr radius. For a hydrogen atom, ao = 0.529 Å. What is the probability of finding an electron within one Bohr radius of the nucleus? normalized probability: 0.323 Why is the probability of finding the electron at the Bohr radius not equal to 1? The electron may exist at a range of radii. The Bohr radius is only the most probable distance of the electron from the nucleus. The electron may exist at a range of radii. The Bohr radius is at the highest probability density. The electron may exist at a range of radii. The Bohr radius is just the radius at which there is equal probability of finding the electron inside the radius as outside. The electron may exist at a range of radii. The Bohr radius is the average distance of the electron from the nucleus. What is the probability of finding an electron of the hydrogen atom within a 1.15a, radius of the hydrogen nucleus? normalized probability:

Answers

a) The normalized probability is

para los – esp (-22) +20.R + 2x) do

Putting R=ao for 1 Bohr radius, we have,

p=519 ao -200 )(a + 21,2, + 20%) = P = [1 – exp(-2)(1+2+2)] P=0.323

Thus, your answer 0.323 is correct

b)

Bohr radius is the most probable distance between electron and the nucleus but the electron can be anywhere. Thus, the answer will be first option which you have selected.

2nd option is incorrect because the highest probability density is at the nucleus. At Bohr radius, it is the highest radial probability density.

3rd option is incorrect because we have calculated in the previous part itself that probability of finding electron is 0.323 and not 0.5

4th option is incorrect because the average distance of electron from nucleus is 1.5 times the Bohr radius.

c)

Here, we put R=1.15ao

Thus, we will get,

do Puzles – exp (2* .) (ač + 2a, x 1.150. + 2 x (1.150.)?)] = P = [1 – exp(-2.30)(1+2.30 + 2.645)] P=0.404

.

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