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## Once a photon of light is generated, how long does it take to escape beyond the outer part of the Sun?

Phillip E.
Featured 2 years ago

It takes thousands of years for a photon to get from the sun's core to escape from the surface.

#### Explanation:

When a photon is created in the core of the sun it will collide with many protons and electrons in its path to the surface.

The photon effectively has to find a path to the surface by randomly being absorbed and re-emitted by particles on the way. This is an example of the drunkard's walk problem.

The number of steps, between particles on the way $n$ can be calculated by the formula $n = {R}^{2} / {d}^{2}$. Where $R = 6.96 \cdot {10}^{8}$m is the distance to travel - radius of the Sun and $d$ is the free mean path - the distance between particles. The value of $d$ is estimated to be about 1 centimetre.

Putting $d = 0.01$ into the formula gives $n = 4.8 \cdot {10}^{21}$ steps.

The time taken for each step is $t = \frac{d}{c}$ where $c = 2.99 \cdot {10}^{8}$m/s is the speed of light. This gives $t = 3.3 \cdot {10}^{- 11}$ seconds.
This gives a total time $T = n t = 1.58 \cdot {10}^{11}$ seconds. Given that a year is $3.1 \cdot {10}^{7}$ seconds. This gives $T = 5109$ years.

This calculation assumes that the free mean step is actually about a centimetre, it could be much smaller, which would make the time longer. In any case it takes thousands of years to make the journey.

## Do black holes defy the laws of physics?

Phillip E.
Featured 2 years ago

Black holes challenge the laws of physics as we know them.

#### Explanation:

Nothing should be able to defy the laws of physics. If something is inconsistent with the laws of physics then they need to be modified to accommodate the inconsistency.

Black holes are extreme objects. They were predicted from the Schwarzschild solution to Einstein's General Theory of Relativity. Many people didn't think that they existed until evidence was found. A black hole could explain the galactic X-ray source at Cygnus X-1. It is now believed that most large galaxies have a supermassive black hole at their centres.

One issue with black holes is that the theories suggest that there is a singularity inside them. A singularity is a point of infinite density and infinite curvature of spacetime. The physicist Kip Thorne described the singularity as the point where all laws of physics break down.

Another problem with black holes is the black hole information paradox. The issue is that if a particle falls into a black hole information about its state is lost. This is forbidden by the laws of physics as we know them. Stephen Hawking is working on a new theory by which the information is somehow retained at the event horizon.

Clearly we need new laws of physics if we are to completely understand black holes. As nothing which goes past the event horizon can ever escape. This makes it impossible to see inside a black hole.

So, yes, black holes defy the laws of physics as we know them. This means that our laws of physics are incomplete.

## Given the following, what is the tension in the string when the nearest mass is a distance of 686 km from the Black hole? #G=6.673×10^-11 m^3kg-1s-2#

Oscar L.
Featured 1 year ago

$3380 \text{ N}$. Don't try this at home: you might get torn apart (see the explanation).

#### Explanation:

The amount of tension is given by the difference in gravitational force between the two masses. Let A and B be the masses, with A being closer to the black hole O. Then:

${F}_{O A} = \frac{G {M}_{O} {M}_{A}}{{r}_{O A}^{2}}$

${F}_{O B} = \frac{G {M}_{O} {M}_{B}}{{r}_{O B}^{2}}$

As the A and B masses are equal we substitute ${M}_{A}$ for ${M}_{B}$ and subtract to get the tension:

$T = {F}_{O A} - {F}_{O B} =$
$G {M}_{O} {M}_{A} \left(\frac{1}{{r}_{O A}^{2}} - \frac{1}{{r}_{O B}^{2}}\right)$

Now put in numbers and calculate:

$G = 6.67 \times {10}^{- 11} {\text{ Nm"^2/"kg}}^{2}$ (same units in SI system as those given)

${M}_{A} \left(= {M}_{B}\right) = 1 \text{ kg}$, taken as exact values

${r}_{A} = 686 , 000 - 0.5 = 685 , 999.5 \text{ m}$

${r}_{B} = 686 , 000 + 0.5 = 686 , 000.5 \text{ m}$

Then

$T = 3380 \text{ N}$. This is over 170 times the gravity of Earth on the same amount of mass. A similar 100-g+ force acting on your mass would pull you apart, even though you are still relatively far from the hole's horizon.

## How much energy is required to release one photon?

Phillip E.
Featured 12 months ago

The energy required to release a photon depends on its frequency.

#### Explanation:

The energy of a photon is related to its frequency or wavelength. The energy $E$ of a photon with frequency $\nu$ is given by the equation:

$E = h \nu$

Where #h = 6.62607004 × 10^-34 kg\ m^2 \/ s# is Planck's constant.

Photons are commonly emitted by electrons in atoms. Electrons can only occupy certain discrete energy levels. When an atom is in an excited state, one or more of its electrons are pushed into a higher energy level. This requires energy. When an electron drops into a lower energy level it releases the energy by releasing a photon at a precise frequency which corresponds to the different between the two energy levels.

Photons can have any frequency from the very low frequency radio waves, through visible light up to the very high frequency gamma rays.

Lower frequency photons are emitted by electrons losing energy. Gamma rays are emitted when an atomic nucleus needs to lose energy.

Hence a photon can be released by any amount of energy.

## How is the sun an example of nuclear energy?

Phillip E.
Featured 6 months ago

The Sun is a good example of nuclear energy as it utilises nuclear fusion, radioactive decay and particle annihilation.

#### Explanation:

The Sun's core is mainly Hydrogen under high temperatures and pressures. The Sun's main source of energy is the proton-proton chain reaction which actually involves three types of nuclear energy producing reactions.

First of all the temperature and pressure allow two protons get close enough for the strong nuclear force to overcome the electrostatic repulsion and fuse them into the highly unstable Helium-2.

#"_1^1H +## "_1^1H->##"_2^2He#

Most of the Helium-2 nuclei fly apart, but relatively rarely the weak force will turn a proton into a neutron a positron and an electron neutrino to for deuterium and releasing energy.

#"_2^2He->##"_1^2H + e^+ + nu_e#

The positron almost immediately annihilates with an electron releasing more energy.

A proton then fuses with deuterium to produce Helium-3 plus a high energy photon.

#"_1^2H##+ "_1^1H##->"_2^3He + gamma#

Finally two Helium-3 nuclei combine to form Helium-4 and two energetic protons.

#"_2^3He##+"_2^3He##->"_2^4He##+2 "_1^1H#

So, there are three nuclear reactions going on in the Sun, fusion, weak decay and electron-positron annihilation.

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