[John369 (OP)]

Would an object having 15 million degree celsius emit gamma rays? And if so, how many gamma rays would it emit per second, given the surface area of the object = 2 meter square. Please show the full calculation, use assumptions wherever necessary.

Please also show the calculation on how to find how much radiation would be emitted from a body at a certain temperature.

[Bored chemist]

It depends; is it something that normally emits gamma rays?
If not, then somewhere between not many, and none.
The Boltzmann constant is about 9 *10^-5 eV per K
so to get an object hot enough to emit gamma rays at, say, 1 MeV you need a temperature approaching ten billion degrees.

very roughly.

[evan_au]

Would an object having 15 million degree celsius emit gamma rays?

The core of the Sun is at around 15 million degrees Celcius, and it produces gamma rays.

This is because the Sun's core consists mostly of Hydrogen, compressed to about 150 times the density of water, and held there for billions of years, allowing Hydrogen fusion to occur.
See: https://en.wikipedia.org/wiki/Solar_core

In this case, the gamma rays come from nuclear fusion, not from black-body radiation (which was what BC calculated).

If the center of the Sun were 100% Helium, there would be no gamma rays emitted, because this temperature is too low to cause Helium fusion. However, the Black-Body radiation is pretty much the same regardless of whether the core is Hydrogen,Helium or Iron.

The diagram here shows that a proton reacting with deuterium produces a gamma ray:
https://en.wikipedia.org/wiki/Stellar_nucleosynthesis#Hydrogen_fusion

[Kryptid]

To calculate the power radiated by a black body with a known temperature, emissivity and surface area, you can use the following equation:

P = AεσT⁴, where

"P" is the resulting power in watts
"A" is the surface area of the black body in square meters
"ε" is the emissivity of the material that the black body is composed of (a number between 0 and 1)
"σ" is the Stefan-Boltzmann constant, which is 5.670373 x 10^-8 W m^-2 K^-4, and
"T" is the temperature of the black body in kelvins

Let's consider a sphere of anodized aluminum (which has an emissivity of 0.9) that has a radius of 0.05 meters and a temperature of 900 kelvins. Using the equation A = 4πr², we determine that a sphere with a radius of 0.05 meters has a surface area of 0.031 square meters:

P = AεσT⁴
P = (0.031)(0.9)(5.670373 x 10^-8)(900⁴)
P = (0.031)(0.9)(5.670373 x 10^-8)(656,100,000,000)
P = (0.0279)((5.670373 x 10^-8)(656,100,000,000)
P = (0.000000001582034067)(656,100,000,000)
P = 1,037.97 watts

If you want to know the peak wavelength of the radiation emitted by a black body, you can use this equation:

λ_max = b/T, where

"λ_max" is the peak wavelength in meters
"b" is Wien's displacement constant, which is 2.8977729 x 10⁻³ m K, and
"T" is the temperature in kelvins

So to find the peak wavelength that the aluminum sphere emits at, we enter the values into the equation:

λ_max = b/T
λ_max = (2.8977729 x 10⁻³)/(900)
λ_max = 3.2197477 x 10^-6 meters (or 3.2197477 micrometers, which is in the infrared)

[John369 (OP)]

Thank you very much.

By Stephen Boltzmann equation, I found that the power emitted is 5e+21 joules per second. And assumed the body wouldn't get evaporated, thought of it as some future technology.

By Wien's Displacement equation, wavelength of emitted radiation = 1.9e-10 meters which lies in X ray region.

But now I want to calculate how many X ray particles per second would be emitted from that object with surface area of 2 meter square. I tried using photon energy equation and got 0.6e-4 eV = 9.6e-24 joules. Considering one X ray photon has that much energy, so 5e+21 joules would have number of X ray photons = 5e+21 / 9.6e-24 = 5e+44 that are emitted per second.

Is the above calculation of number of X ray photons emitted per second correct? Please notify. 

[Kryptid]

Is the above calculation of number of X ray photons emitted per second correct?

The problem is that not all of the photons emitted are X-ray photons. There will also be a significant number of photons emitted in the ultraviolet, visible, infrared and microwave frequencies. Given that the spread of those photons follow something of a bell curve, you might need calculus in order to find out exactly what percentage of those photons fall in the X-ray frequencies. I am awful at calculus, unfortunately, and thus cannot help there.

[Bored chemist]

The maths you want is here
https://en.wikipedia.org/wiki/Planck%27s_law
Good luck.
The answer is still going to be practically zero.

[evan_au]

anodized aluminum (which has an emissivity of 0.9)

That is fine for 900C.

For 15 million C, all the outer electrons are ripped off and it becomes a plasma; room-temperature emissivity of a solid metal does not apply.

At these temperatures, most of the radiation is bremsstrahlung radiation from freely-moving electrons passing near a freely-moving nucleus (plus a rare gamma ray from a Hydrogen fusion event).

See: https://en.wikipedia.org/wiki/Bremsstrahlung

[Kryptid]

For 15 million C, all the outer electrons are ripped off and it becomes a plasma; room-temperature emissivity of a solid metal does not apply.

I figured so. Any idea if the emissivity of such plasma has ever been measured?

[evan_au]

I figured so. Any idea if the emissivity of such plasma has ever been measured?

I am sure fusion researchers have measured it...

See: https://en.wikipedia.org/wiki/Bremsstrahlung#Bremsstrahlung_in_plasma

[Bored chemist]

For 15 million C, all the outer electrons are ripped off and it becomes a plasma; room-temperature emissivity of a solid metal does not apply.

I figured so. Any idea if the emissivity of such plasma has ever been measured?

I'm fairly sure that a sufficiently hot plasma is a black body.
It consists of essentially free ions and electrons and those can have any value of kinetic energy so there's no "spectrum" of absorption. The incoming photons just throw the electrons about faster.
And the interaction of a free electron or ion with a photon is a spectroscopically allowed transition.

In addition, if the plasma is "dense" enough then collisional broadening will wash out any spectral detail leaving a "black" object.

[Colin2B]

......Any idea if the emissivity of such plasma has ever been measured?

This paper says “We determined that all the laser-induced air plasma spectra are from a continuous spectrum and some line spectra superposed on the continuous spectrum. The stronger short wavelength continuous spectrum is caused by bremsstrahlung radiation of electrons in the plasma, and the weaker long wavelength continuous spectrum is caused by electron and ion recombination. Line spectra originate from excited molecules, atoms, and their first-order ions, but no line spectra form higher-order ions.” https://www.osapublishing.org/ao/abstract.cfm?uri=ao-49-13-C80

The OP might also be interested in Sonoluminescence, where acoustic waves in water cause bubbles which emit light. Putterman of UCLA introduced a theory known as shock wave SL. Shock wave theory suggests that as the bubble collapses, it retains its spherical shape creating a high pressure system within the bubble. As the bubble nears its minimum radius, the Van Der Waals force become significantly large and almost instantly stops the bubble from collapsing any further. This fast deceleration of the bubble walls can exceed 107 times that of gravity. Even though the walls have stopped, a shock wave continues to travel towards the center of the bubble. The energy contained in the shock wave then becomes focused at the centralized point of the bubble and creates an area of extremely high temperatures. These temperatures result in the production of plasma and begin to emit photons in a process known as thermal Bremsstrahlung. The photons emitted from this process are over a broad spectrum.
See also https://journals.jps.jp/doi/abs/10.1143/JPSJ.69.112
“A mechanism of light emission from a single bubble sonoluminescence (SBSL) was formulated by assuming that the source for the light emission is bremsstrahlung in partially ionized gases. This emission mechanism turns out to be same as the black body radiation with finite absorption, which confirms that the principal mechanism of SBSL is bremsstrahlung with slight black body emission nature.”

[Bored chemist]

These temperatures result in the production of plasma and begin to emit photons in a process known as thermal Bremsstrahlung.

That's a more sciency way to say essentially this

The incoming photons just throw the electrons about faster.

But looked at from the other perspective- emission vs absorption.

[yor_on]

I f*ng love you guys.
Keep on writing.

[Colin2B]

That's a more sciency way to say essentially this

The incoming photons just throw the electrons about faster.

True, but then they are writing a paper for publication so they gotta sound posh.
But at least it has some of the measurements @Kryptid was asking about 

[evan_au]

The OP may also be interested in gamma ray emissions above lightning strikes. The atmosphere is nowhere near millions of degrees (on average).

The gamma rays have been detected by satellites, and are thought to originate in the upper atmosphere, perhaps by electrons accelerated to relativistic speeds by the voltages gradients in the upper atmosphere. This occurs when lightning suddenly discharges a cloud lower in the atmosphere.

In at least one case, the spectrum suggested that the event had produced positrons, which had then annihilated with electrons in the upper atmosphere, producing gamma rays with a characteristic energy.

See: https://en.wikipedia.org/wiki/Terrestrial_gamma-ray_flash

[evan_au]

Even measuring a temperature of millions of degrees is a challenge.

This nuclear fusion research reactor at the Max Plank Institute in Germany has operated at a temperature of 40 million degrees. Most of the radiation follows a black-body spectrum, with peak emitted energy is at X-Ray frequencies.

The experimenters measure the temperature of the nuclei and the electrons independently:
- They are not necessarily the same, especially while the plasma is being heated
- The electron temperature is measured by firing a laser through the plasma. They can calculate the electron density at different points by measuring the speed of light in the plasma, and measure its temperature at different points by seeing how the frequency of scattered laser light is broadened by Doppler shift.
- At these temperatures, Hydrogen & Helium ions are fully ionised, and so don't have a line spectrum. They measure the ion temperature indirectly by looking at the emission spectrum of heavier impurity atoms in the plasma. Apparently, these heavy ions are not totally ionised in this plasma, and so they emit a line spectrum after collision with lighter nuclei; doppler broadening of this line spectrum can be detected and measured.
- They need to minimise the amount of heavy ions in the plasma, as they radiate energy out of the plasma - but small quantities are useful for monitoring the temperature.

A discussion of temperature measurement starts at 3:00:00: https://omegataupodcast.net/312-the-wendelstein-7-x-fusion-experiment/

[Bored chemist]

Even measuring a temperature of millions of degrees is a challenge.

Even defining a temperature of millions of degrees  is a challenge.
The electronic, nuclear, translational, vibrational and rotational temperatures may all be different (though, at those temperatures there's not much left to vibrate or rotate)