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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>
Answer:
Work done by battery will be equal to 10.238 J
Explanation:
We have given ideal an ideal battery with ideal voltage V = 24 volt
Constant current
Resistance will be equal to
Time is given t = 3 minutes = 3×60 = 180 sec
We have to find work done by battery
Work done is equal to
So work done by battery will be equal to 10.238 J
Answer:
Explanation:
Given
Sphere A has a net charge of +5 Q
Sphere B has a net charge of -3 Q
Sphere C has a net charge of 7 Q
When sphere B is touched to sphere A then charge is redistributed among the sphere such that charge on each sphere is
Now sphere A and B has charge Q
When sphere A is touched to sphere C
Net charge after removal
Now sphere A and C has 4 Q charge
Now sphere C and B is touched
Net charge on C and B is given by
At the end of process charge on sphere A is 4Q
Charge on B and C is 2.5Q
Answer:
Option (1), option (4) and option (5)
Explanation:
The main observations of Ernest Rutherford's experiment are given below:
1. most of the positively charged particles pass straight, it means there is an empty space in the atom.
2. Very few positively charged particles retraces their path.
So,
The positively charged particles were deflected because like charges repel, that means they are deflected by protons.
Almost all the positively charge concentrate in a very small part which is called nucleus.