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## During the fusion process, how is mass converted into energy?

Waleed A.
Featured 1 year ago

$E = m {c}^{2}$

#### Explanation:

This is calculated using the famous equation of Einstein,

$E = m {c}^{2}$

In Fusion reaction like the ones taking place in the core of a Star, there is enough pressure to fuse hydrogen nuclei to form one helium nucleus.

So, 4 hydrogen nuclei are fused together to form one Helium nucleus. But, where does the energy come from that keeps the Sun from collapsing?.

When 4 Hydrogen nuclei are merged together they show a certain discrepancy in the mass when a Helium atom is formed, i.e the mass of 4 Hydrogen atoms before Fusion is higher than the mass of the Helium atom after the reaction this mass defect is converted into energy by $E = m {c}^{2}$.

$\text{mass of a hydrogen atom " = " 1.00794 u}$

$\text{mass of one helium atom " = " 4.002602 u}$

Where $u = 1.6605 \times {10}^{- 27} \text{kg}$.

The mass defect, $\Delta m$, will be

$\Delta m = 4 \times {m}_{H} - {m}_{H e}$

$= 4 \times \text{1.00794 u" - "4.002602 u}$

$= \text{0.029158 u }$ or $\text{ "4.8416859 * 10^(-29)"kg}$

In the equation $E = m {c}^{2}$, $c$ is the speed of light in a vacuum, approximately equal to $3 \cdot {10}^{8} {\text{m s}}^{- 1}$.

This means that you have

$E = 4.8416859 \cdot {10}^{- 29} {\text{kg" * (3 * 10^8)^2"m"^2"s}}^{- 2}$

$E = 4.35751731 \cdot {10}^{- 12} \text{ J per reaction}$

## What is the sun's atmosphere called?

Hunaid L. Hanfee
Featured 1 year ago

Just as Earth, the Sun's atmosphere is divided in 3 parts.
$1.$ Photosphere
$2.$ Chromosphere
$3.$ Corona

#### Explanation:

$1.$ Photosphere - The lowest and deepest layer of the Sun's atmosphere is the photosphere. This layer is about 400 kilometers (250 miles) thick. The temperature in the photosphere varies between about 6500 K (${6200}^{o} c$) at the bottom and 4000 K ($3700 c$) at the top. In this layer the Sun's energy is released as light. Solar flares are formed in this layer.

$2.$ Chromosphere - The chromosphere is a layer about 400 km (250 miles) and 2100 km (1300 miles) above the photosphere. The temperature in the chromosphere varies between about 4000 K (${3700}^{o} c$) at the bottom and 8000 K (${7700}^{o} c$) at the top. The chromosphere emits a reddish glow as super-heated hydrogen burns off. But the red rim can only be seen during a total solar eclipse. This light can be seen only in total solar eclipse because it weaker than the light of other part.

$3.$ Corona - This layer starts from 2100 km ( 1300 miles) above the photosphere. The temperature in the corona is 500,000 K ($500 , {000}^{o} c$). The corona cannot be seen with naked eye except during a total eclipse of the Sun.

Here, I am uploading a picture by which you can understand it much better.

## In a star if helium fuses to make carbon, what makes the elements in between?

Phillip E.
Featured 1 year ago

The elements Lithium, Beryllium and Boron are made by cosmic rays.

#### Explanation:

The elements Hydrogen and Helium were made during the Big Bang. Helium is also made by fusion reactions in stars.

The elements Lithium Beryllium and Boron can't be made in any quantity is stars as they are intermediate steps in other fusion reactions. Any of these elements present at the birth of a star will get consumed by later reactions.

These intermediate elements (Lithium, Beryllium and Boron) are produced by cosmic rays. These are high energy protons and other particles. These collide with nuclei of Carbon and Oxygen which breaks them down and produces Lithium, Beryllium and Boron.

Carbon is produced by reactions such as the triple Helium reaction. This reaction explains what happens to Beryllium.

${\text{_2^4He+""_2^4He rarr }}_{4}^{8} B e$
${\text{_4^8Be+""_2^4He rarr }}_{6}^{12} C$

The diagram shows where all of the elements are created.

## Does the Oort cloud rotate? Or is it static?

Parzival S.
Featured 1 year ago

The Oort Cloud certainly moves and most probably revolves around the Sun.

#### Explanation:

Does the Oort cloud rotate? Excellent question!

Ok - first thing to know is that the Oort cloud is thought to exist but hasn't been proven to exist - so that's the first thing to keep in mind.

If in fact the Cloud does exist, the next question is - does it move? The answer is almost certainly yes. Keep in mind that in deep space there are 2 factors working out there:

• gravity - so the pull of one object by another, whether it's elements of the Cloud on other elements, the Sun and other planets pulling on it, and the gravitational effects of neighbouring stars, and

• the lack of friction to stop the movement of the objects being pulled.

So then the question becomes - does the Oort Cloud revolve around the Sun (to rotate means to spin on it's own axis, such as the Earth as it spins and causes the Sun to rise and set. To revolve is to move around a different object, such as the Earth as it moves around the Sun and creates seasons)?

Again, the answer is almost certainly yes. Keep in mind that it's the gravitational pull of the Sun on these objects that keeps them from simply floating off into deep space, so there is a constant pull from the Sun. But they also aren't falling into the Sun, and so the motion of the objects of the Cloud, and therefore the Cloud itself, would orbit the Sun.

As an example we are more familiar with, let's look to Saturn's rings. The rings themselves revolve around Saturn and the speed is dependent on how far away the ring is from Saturn's surface.

http://spacemath.gsfc.nasa.gov/weekly/10Page28.pdf

## Mars has an average surface temperature of about 200K. Pluto has an average surface temperature of about 40K. Which planet emits more energy per square meter of surface area per second? By a factor of how much?

JMicheli
Featured 1 year ago

Mars emits $625$ times more energy per unit of surface area than Pluto does.

#### Explanation:

It is obvious that a hotter object will emit more black body radiation. Thus, we already know that Mars will emit more energy than Pluto. The only question is by how much.

This problem requires evaluating the energy of the black body radiation emitted by both planets. This energy is described as a function of temperature and the frequency being emitted:

$E \left(\nu , T\right) = \frac{2 {\pi}^{2} \nu}{c} \frac{h \nu}{{e}^{\frac{h \nu}{k T}} - 1}$

Integrating over frequency gives the total power per unit area as a function of temperature:
${\int}_{0}^{\infty} E \left(\nu , T\right) = \frac{{\pi}^{2} c {\left(k T\right)}^{4}}{60 {\left(\overline{h} c\right)}^{3}}$

(note that the above equation uses $\overline{h}$, the reduced Planck's constant, rather than $h$. It is difficult to read in Socratic's notation)

Solving for the ratio between the two, then, the result is incredibly simple. If ${T}_{p}$ is Pluto's temperature and ${T}_{m}$ is Mars' temperature then the factor $a$ can be calculated with:

$\frac{{\pi}^{2} c {\left(k {T}_{m}\right)}^{4}}{60 {\left(\overline{h} c\right)}^{3}} = a \frac{{\pi}^{2} c {\left(k {T}_{p}\right)}^{4}}{60 {\left(\overline{h} c\right)}^{3}}$
$\frac{\cancel{{\pi}^{2} c {k}^{4}}}{\cancel{60 {\left(\overline{h} c\right)}^{3}}} {T}_{m}^{4} = a \frac{\cancel{{\pi}^{2} c {k}^{4}}}{\cancel{60 {\left(\overline{h} c\right)}^{3}}} {T}_{p}^{4}$
${\left({T}_{m} / {T}_{p}\right)}^{4} = a = {\left(\frac{200}{40}\right)}^{4} = {5}^{4} = 625$ times as much

## What determines the amount of gravitational contraction in a star?

Phillip E.
Featured 3 months ago

All bodies need to reach an equilibrium with gravity to stop further collapse.

#### Explanation:

Most small bodies, which includes comets and most asteroids, have insufficient gravity overcome the rigidity of their component materials to undergo collapse. This means that they can be any shape.

Larger bodies, such as most moons, all planets and all stars, have sufficient gravity to overcome material rigidity and undergo a level of collapse. This is why these bodies are almost spherical in shape.

Almost spherical bodies are in hydrostatic equilibrium where gravity is balanced out by internal pressure to prevent further collapse.

In the case of main sequence stars, the outward pressure generated by fusion reactions in the core keep the star at a relatively large size. All main sequence stars are in hydrostatic equilibrium.

When a star runs out of fuel for fusion reactions, then gravitational collapse of the core is inevitable.

If the stellar core is less than about 1.4 solar masses then gravitational collapse is stopped by electron degeneracy pressure. This is a quantum effect caused by the Pauli exclusion principle. No two electrons can be in the quantum state. This prevents atoms from having overlaying electron shells. The most degenerate state this allows is the white dwarf star.

If the stellar core exceeds 1.4 solar masses then electron degeneracy pressure fails and protons and electrons are forced to combine into neutrons forming a neutron star. This generates a huge supernova explosion in the outer layers of the star.

Neutron stars are held in hydrostatic equilibrium by neutron degeneracy pressure. This is another quantum effect which prevents two neutrons being in the same quantum state.

If the neutron star is more than about three solar masses then gravity overcomes it. This means that gravity can now collapse the stellar core to its extreme which is a black hole.

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