Suppose you built a scale-model atom in which the nucleus is the size of a tennis ball. About how far would the cloud of electrons extend?

Answers

Answer 1
Answer:

Answer:

500 m

Explanation:

The size of a nucleus is about

d=10^(-15) m

Instead, the size of the electron cloud extends at the order of

D=10^(-10)m

This means that the ratio between the size of the electron cloud and the size of the nucleus is about

k=(D)/(d)=10,000

So, the electron cloud is about 10,000 bigger than the nucleus of the atom.

Here we want to build a scale-model atom, in which the nucleus is the size of a tennis ball, so a size of approximately

d=5 cm

Therefore, since the proportions must be respected, the electron clouds must be 10,000 bigger, so its size in this model would be:

D=kd=(10,000)(5)=50,000 cm = 500 m

So, the size of the electron cloud would be 500 m.

Answer 2
Answer:

Final answer:

The cloud of electrons in a scale-model atom would extend about 100 meters.

Explanation:

The nucleus is a central, positively charged region within an atom. It contains protons, which have a positive electric charge, and neutrons, which are electrically neutral. Nuclei make up the majority of an atom's mass and are held together by the strong nuclear force, while electrons orbit the nucleus in electron shells.

In a scale-model atom where the nucleus is the size of a tennis ball, the cloud of electrons would extend about 100 meters. This can be calculated by comparing the size of the nucleus to the actual size of an atom. The diameter of a tennis ball is approximately 6.7 cm, while the diameter of an atom is on the order of 0.1 nm. By scaling up, we find that the electron cloud would extend around 100 meters.

Learn more about Scale-model atom here:

brainly.com/question/33541403

#SPJ12


Related Questions

What has to be different to allow refraction to occur?a. the medium b. the light c. the wave d. the speed
A satellite with a mass of 5.6 E 5 kg is orbiting the Earth in a circular path. Determine the satellite's velocity if it is orbiting at a distance of 6.8 E 5 m above the Earth's surface. Earth's mass = 5.98 E 24 kg; Earth's radius = 6.357 E 6 m.A) 6,800 m/sB) 7,200 m/sC) 7,500 m/sD) 7,900 m/s
A crate with a mass of 3.00 kg starts from rest at the top of a 27.0∘ incline and slides 2.00 m down the incline in 1.60 s. (a) What is the magnitude of the acceleration of the crate (in m/s² )? (b) What is the frictional force (in N) acting on the crate? (Enter the magnitude.) (c) What is the coefficient of kinetic friction between the crate and the incline? (d) What is the speed of the crate (in m/s ) after it has slid 2.00 m ?
two vectors of same units have magnitude of 8 unit and 5 unit. what are the maximum and minimum magnitude of resultant that can be obtained with the two vectors?
Select the correct answer. Rita is a registered dietician. What does her work entail? A. prescribing medication for clients B. cooking healthy meals for students C. demonstrating how to use gym equipment D. making recommendations for healthy eating habits

What are the disadvantages of driverless cars? Check all that apply

Answers

They cannot tell the difference between people and other objects.

They will wait for crossing pedestrians even if signaled to keep driving. Explanation:

Just did it

Answer:

Explanation:

,,,,,,,,,,,,,,,,,,,,,,,,,,,

A 0.4-kg toy train car moving forward at 3 m/s collides with and sticks to a 0.8–kg toy car that is traveling in the opposite direction at –2 m/s. What is the size and direction of the final velocity of the two cars

Answers

Hey there!

Seems like you're looking for the size and direction to the final velocity of the two cars. To find it, you must solve it like this.

0.4 kg(3 m/s) + 0.8kg(–2 m/s) = 1.2 kg m/s -1.6 kg m/s = –0.4 kg m/s

–0.4 kg m/s = 1.2 kg(v) = (–0.4 kg m/s)/(1.2 kg) = v = –0.33 m/s


So, the cars are traveling at -0.33 m/s in the direction of the second car.


Hope this helps


Tobey

Within the metric system all units are based on the number?

Answers

The metric system is based upon powers of ten (10), which is convenient because: A measurement in the metric system that is represented by a rational number remains a rational number after metric unit conversion.

Answer:

10

Explanation:


Why is gravity an example of scientific law?

Answers

Gravity can be an example of scientific law because of the fact that it has been proven over and over, again and again. While a theory is proven by evidence, it can be disproved. Laws rarely change. Although there are theories about gravity as well. The difference is a theory explains why it happens while a law lets us calculate a way to find what happened. In Newton's Law of Universal Gravitation, the formula will allow you to calculate the gravitational pull of the Earth and what ever you dropped for instance. However Einstein's Theory of General Relativity explains why whatever you dropped fell. So it depends what you mean by gravity being a law. It can be both depending on if your using it to prove something or to explain why that force occurs. :)

Because it is a basic concept that defines a constant and universal behaviour of nature, gravity is seen as an illustration of a scientific law.

Scientific laws are succinct formulations that precisely explain a natural occurrence and are the result of in-depth observation and investigation. Einstein's theory of general relativity or Newton's law of universal gravitation both use the concept of gravity to explain how mass-containing objects interact with one another in space.

It offers a mathematical framework that enables researchers to perform precise projections and estimates about celestial body motion, the behaviour of things close to the surface of the Earth, and the structure of the cosmos.

The fact that gravity is recognised as a scientific rule is a reflection of the fact that it is widely accepted and has been shown to have practical applications.

To know more about gravity :

brainly.com/question/31321801

#SPJ6.

After generating a hypothesis, a scientist next: A. designs a series of tests. B. formulates a theory. C. makes a prediction. D. writes a grant proposal. E. does an experiment.

Answers

The answer is (c) because

A drowsy cat spots a flowerpot that sails first up and then down past an open window. The pot is in view for a total of 0.66 s, and the top-to-bottom height of the window is 1.80 m. How high above the window top does the flowerpot go? (Assume that in both sailing upward and then falling downward, the flowerpot is seen to span the full vertical height of the window.)

Answers

Answer:

Height covered after crossing window top = 0.75 m

Explanation:

In the question,

The height of the window pane = 1.8 m

Time for which the flowerpot is in view = 0.66 s

So,

The time for which it was in view while going up is = 0.33 s

Time for which it was in view while going down = 0.33 s

So,

Let us say,

The initial velocity of the flowerpot = u m/s

So,

Using the equation of the motion,

s=ut+(1)/(2)at^(2)\n1.8=u(0.33)-(1)/(2)(9.8)(0.33)^(2)\n1.8=0.33u-0.5336\n0.33u=2.33\nu=7.07\,m/s

So,

Velocity at the top of the window pane is given by,

v=u+at\nv=7.07-(9.8)(0.33)\nv=3.836\,m/s

Now,

Let us say the height to which the flowerpot goes after crossing the window pane is = h

So,

Using the equation of motion,

v^(2)-u^(2)=2as\n(0)^(2)-(3.836)^(2)=2(-9.8)h\nh=(14.714)/(19.6)\nh=0.75\,m

Therefore, the height covered by the flowerpot after window is = 0.75 m