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vagabundo [1.1K]
2 years ago
7

Consider an electron within the 1s orbital of a hydrogen atom. the normalized probability of finding the electron within a spher

e of a radius r centered at the nucleus is given by
Physics
2 answers:
I am Lyosha [343]2 years ago
4 0
Thank you for posting your question here at brainly. I think your question is incomplete. Below is the complete question, it can be found elsewhere:

What is the probability of finding an electron within one Bohr radius of the nucleus?<span>Consider an electron within the 1s 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 1-a0^2[a0^2-e^(-2R/a0)(a0^2+2a0R+2R2)]. Where a0 is the Bohr radius (for a hydrogen atom, a0 = 0.529 Å.). What is the probability of finding an electron within one Bohr radius of the nucleus? What is the probability of finding an electron of the hydrogen atom within a 2.30a0 radius of the hydrogen nucleus?

Below is the answer:

</span><span>you plug the values for A0 and R into your formula</span>
zzz [600]2 years ago
4 0

Answer:

the normalized probability of finding the electron within sphere of a radius r centered at the nucleus is given by:

Normalized probability =  1/ a_{0} ^2[a_{0} ^2-e^2^r^/^a_{}  (a_{0} ^2+2a_{0} r+2r^2)]

where a_{0} is Bohr radius ( for a hydrogen atom a_{0}  =0.529 A

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Answer:

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Initial kinetic energy of white ball is Vi1 = 3.09m/s

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Vf2 = 3.09m/s

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