Homework Answers
a) The normalized probability is
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](http://img.homeworklib.com/questions/b6d0c1c0-700f-11ea-8161-f5a3b0b61c88.png?x-oss-process=image/resize,w_560)
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](http://img.homeworklib.com/questions/b73083f0-700f-11ea-9355-f9d55e792799.png?x-oss-process=image/resize,w_560)
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In a one electron system, the probability of finding the
electron within a shell of thickness δr at a radius of r from the
nucleus is given by the radial distribution function,
P(r)=r2R2(r).
An electron in a 1s hydrogen orbital has the radial wavefunction
R(r) given by
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where a0 is the Bohr radius (52.9 pm).
Calculate the probability of finding the electron in a sphere of
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