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bekas [8.4K]
3 years ago
13

How does fission and fusion affect an atoms nucleus differently?

Physics
1 answer:
Nata [24]3 years ago
5 0

When a heavy nucleus breaks up into two or more fragments with the liberation of energy, the process is called fission.

When two light nuclei fuse together to form a comparatively heavy nucleus, the process is called fusion. Both processes are accompanied by release of energy.

The most common example of fission is the fission of Uranium nucleus, which absorbs a slow neutron and breaks into Barium, Krypton, a few neutrons and energy.

U(235)+neutron→Ba(139)+Kr(94)+3 neutrons+200 MeV

Each fission releases around 200 MeV of energy.

The fusion reaction is responsible for energy generated by stars.

Protons in the stars combine to form a Helium nucleus with the liberation of energy.

4 protons→Helium+2 positrons+ nuetirnos+ 26.7 MeV

Fusion reactions can happen only at very high temperatures, which is required ,so that the protons have enough kinetic energy to overcome their mutual electrostatic repulsion. The temperatures that can result in fusion are of the order 10^8 K is present in stars , hence stars generate energy by fusion.

However, such temperatures are difficult to maintain and control on Earth, and hence controlled fusion reaction is still not available commercially. Experiments are on , using large magnetic fields to contain the ions that undergo fusion, since no container can hold the nuclei while they undergo fusion.

Fusion is also called clean energy, since the byproducts of fusion are not radioactive, but many fission byproducts are generally radioactive.

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The nail has become a temporary magnet, while the

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Item 8 The cost (in dollars) of making b bracelets is represented by 4 5b. The cost (in dollars) of making b necklaces is repres
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Answer:

3b+2

Explanation:

First your question has so many mistakes

What I understand

bracelets cost 4+5b

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First we have to make an equaion

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3 years ago
Rutherford tracked the motion of tiny, positively charged particles shot through a thin sheet of gold foil. Some particles
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Question

Rutherford tracked the motion of tiny, positively charged particles shot through a thin sheet of gold foil. Some particles  travelled in a straight line and some were deflected at different angles.

Which statement best describes what Rutherford concluded from the motion of the particles?

A) Some particles travelled through empty spaces between atoms and some particles were deflected by electrons.

B) Some particles travelled through empty parts of the atom and some particles were deflected by electrons.

C) Some particles travelled through empty spaces between atoms and some particles were deflected by small areas of high-density positive charge in atoms.

D) Some particles travelled through empty parts of the atom and some particles were deflected by small areas of high-density positive charge in atoms.    

Answer:

 

The right answer is C)    

Explanation:

In the experiment described above, a piece of gold foil was hit with alpha particles, which have a positive charge. Alpha particles <em>α</em> were used because, if the nucleus was positive, then it would deflect the positive particles. The principles of physics posit that electric charges of the same orientation repel.  

So most as expected some of the alpha particles went right through meaning that the gold atoms comprised mostly empty space except the areas that were with a dense population of positive charges. This area became known as the "nucleus".  

Due to the presence of the positive charges in the nucleus, some particles had their paths bent at large angles others were deflected backwards.

Cheers!

6 0
3 years ago
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The electric potential in a particular region of space varies only as a function of y-position and is given by the function V(y)
nikdorinn [45]

Answer:

E = 55.9583\ Volts/meter

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V(22.5) = 1.69(22.5)^2 + 15.6*22.5 + 52.5

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Then, to find the magnitude of the electric field, we just need to divide the electric potential by the distance y:

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3 0
3 years ago
A long hollow cylindrical conductor (inner radius = 2.0 mm, outer radius = 4.0 mm) carries a current of 12 a distributed uniform
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Here we can use ampere'a law to find the magnetic field

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B = 4\pi * 10^{-7} (12 + 5)

B = 1.13 * 10^{-3} T

3 0
3 years ago
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