This means that we shouldn't imagine electrons as single objects going around the atom. Instead, all we know is the probability of finding an electron at a particular location. What we end up with is something called an electron cloud. An electron cloud is an area of space in which an electron is likely to be found. It's like a 3-D graph showing the probability of finding the electron at each location in space. Quantum mechanics also tells us that a particle has certain numbers (called quantum numbers) that represent its properties. Just like how materials can be hard or soft, shiny or dull, particles have numbers to describe the properties. These include a particle's orbital quantum numbers, magnetic quantum number, and its spin. No two electrons in an atom can have exactly the same quantum numbers. Orbital quantum numbers tell you what energy level the electron is in. In the Bohr model, this represents how high the orbit is above the nucleus; higher orbits have more energy. The first orbit is n=1, the second is n=2, and so on. The magnetic quantum number is just a number that represents which direction the electron is pointing. The other important quantum mechanical property, called spin, is related to the fact that electrons come in pairs. In each pair, one electron spins one way (with a spin of one half), and the other electron spins the other way (with a spin of negative one half). Two electrons with the same spin cannot exist as a pair. This might seem kind of random, but it has effects in terms of how magnetic material is. Materials that have unpaired electrons are more likely to be magnetic
Protons and neutrons are located in the nucleus, a dense central core in the middle of the atom, while the electrons are located outside the nucleus.
The electric potential V(z) on the z-axis is : V = 
The magnitude of the electric field on the z axis is : E = kб 2
( 1 - [z / √(z² + a² ) ] )
<u>Given data :</u>
V(z) =2kQ / a²(v(a² + z²) ) -z
<h3>Determine the electric potential V(z) on the z axis and magnitude of the electric field</h3>
Considering a disk with radius R
Charge = dq
Also the distance from the edge to the point on the z-axis = √ [R² + z²].
The surface charge density of the disk ( б ) = dq / dA
Small element charge dq = б( 2πR ) dr
dV 
----- ( 1 )
Integrating equation ( 1 ) over for full radius of a
∫dv = 
V = ![\pi k\alpha [ (a^2+z^2)^\frac{1}{2} -z ]](https://tex.z-dn.net/?f=%5Cpi%20k%5Calpha%20%5B%20%28a%5E2%2Bz%5E2%29%5E%5Cfrac%7B1%7D%7B2%7D%20-z%20%5D)
= ![\pi k (\frac{Q}{\pi \alpha ^2})[(a^2 +z^2)^{\frac{1}{2} } -z ]](https://tex.z-dn.net/?f=%5Cpi%20k%20%28%5Cfrac%7BQ%7D%7B%5Cpi%20%5Calpha%20%5E2%7D%29%5B%28a%5E2%20%2Bz%5E2%29%5E%7B%5Cfrac%7B1%7D%7B2%7D%20%7D%20%20-z%20%5D)
Therefore the electric potential V(z) =
Also
The magnitude of the electric field on the z axis is : E = kб 2( 1 - [z / √(z² + a² ) ] )
Hence we can conclude that the answers to your question are as listed above.
Learn more about electric potential : brainly.com/question/25923373
You have to do the math of each and see which one adds up to 66.5
Antoine-Laurent Lavoisier was the first person to report the four element classification system but also ended up including some compounds rather than elements.
