I learnt that in astrophysical spectroscopy, the emission spectrum of distant stars is used to determine what they're made of. So why is it that our own Sun is emitting the whole spectrum ?

Autumn Pham
2022-05-10
Answered

I learnt that in astrophysical spectroscopy, the emission spectrum of distant stars is used to determine what they're made of. So why is it that our own Sun is emitting the whole spectrum ?

You can still ask an expert for help

Sage Guerra

Answered 2022-05-11
Author has **10** answers

There are continuous spectra, emission spectra, and absorption spectra. Dense materials, such as the deeper parts of our sun, have continuous spectra. Hot, low-density gases have emission spectra (a black spectrum with bright lines in it). The continuous spectrum generated deep in our sun passes through the cooler and lower-density outer regions of the sun, where it is partially absorbed. The result is an absorption spectrum: a rainbow with certain dark lines notched out of it

A good way to think about this is that a line spectrum comes from individual atoms, which are quantum-mechanical. In systems with a very large number of particles, the quantum-mechanical behavior becomes undetectable. A dense gas, such as the deeper parts of the sun, has all its atoms so close together that they act like one big object with a very large number of particles. Therefore you get no quantum-mechanical features, and it's just a continuous spectrum.

A good way to think about this is that a line spectrum comes from individual atoms, which are quantum-mechanical. In systems with a very large number of particles, the quantum-mechanical behavior becomes undetectable. A dense gas, such as the deeper parts of the sun, has all its atoms so close together that they act like one big object with a very large number of particles. Therefore you get no quantum-mechanical features, and it's just a continuous spectrum.

asked 2022-05-08

The wavelength of emitted radiation when an electron jumps orbits in the Bohr atomic model is given by

1/$\lambda $ = ${R}_{H}$ ($\frac{1}{{n}_{f}^{2}}$ - $\frac{1}{{n}_{i}^{2}}$) ${Z}^{2}$

But that of X-Ray emission is given by

1/$\lambda $ = $1.1\times {10}^{7}$ ($\frac{1}{{n}_{f}^{2}}$ - $\frac{1}{{n}_{i}^{2}}$) $(Z-1{)}^{2}$

Why is there a difference? Why aren't X-Rays treated like any other emissions?

1/$\lambda $ = ${R}_{H}$ ($\frac{1}{{n}_{f}^{2}}$ - $\frac{1}{{n}_{i}^{2}}$) ${Z}^{2}$

But that of X-Ray emission is given by

1/$\lambda $ = $1.1\times {10}^{7}$ ($\frac{1}{{n}_{f}^{2}}$ - $\frac{1}{{n}_{i}^{2}}$) $(Z-1{)}^{2}$

Why is there a difference? Why aren't X-Rays treated like any other emissions?

asked 2022-05-13

How much does the gravitational redshift change a neutron star emission spectra disturbing so the measurement of its surface temperature?

asked 2022-05-13

I need to be able to convert an arbitrary emission spectrum in the visible spectrum range (i.e. for every wavelength between 380 and 780, I have a number between 0 and 1 that represents the "intensity" or dominance of that wavelength), and I need to be able to map any given spectrum into a particular color space (for now I need RGB or CIE-XYZ). Is it possible?

For the spectrum say I have the emission spectrum of a white light, then every wavelength in the spectrum will have an intensity of 1, whereas for a green-bluish light I'd have most of the wavelengths between 500 and 550 with an intensity close to 1, with other wavelengths gradually dropping in intensity. So the first spectrum should be converted to pure white whereas the other one would be converted to a green-bluish color in any color space.

Is there a way to do this?

For the spectrum say I have the emission spectrum of a white light, then every wavelength in the spectrum will have an intensity of 1, whereas for a green-bluish light I'd have most of the wavelengths between 500 and 550 with an intensity close to 1, with other wavelengths gradually dropping in intensity. So the first spectrum should be converted to pure white whereas the other one would be converted to a green-bluish color in any color space.

Is there a way to do this?

asked 2022-05-08

From my quick investigation, the spectrum is based on the Rydberg formula, and with a small change, would lead to

$\frac{1}{{\lambda}_{\mu}}=\frac{{m}_{\mu}}{{m}_{e}}\left(R(\frac{1}{{n}_{1}^{2}}-\frac{1}{{n}_{2}^{2}})\right)$

where ${m}_{\mu}$ is the mass of a muon.

So, taking hydrogen as an example, we would observe similar bands, shifted into the x-ray/gamma range.

Is this correct?

$\frac{1}{{\lambda}_{\mu}}=\frac{{m}_{\mu}}{{m}_{e}}\left(R(\frac{1}{{n}_{1}^{2}}-\frac{1}{{n}_{2}^{2}})\right)$

where ${m}_{\mu}$ is the mass of a muon.

So, taking hydrogen as an example, we would observe similar bands, shifted into the x-ray/gamma range.

Is this correct?

asked 2022-05-19

Give the reasons of:

1. In a cavity laser, a beam could be transmitted through the cell without being distorted by the cell wall.

2. The statistical weight for most levels are essentially the in dense material.

3. Homogeneous broading.

1. In a cavity laser, a beam could be transmitted through the cell without being distorted by the cell wall.

2. The statistical weight for most levels are essentially the in dense material.

3. Homogeneous broading.

asked 2022-05-10

When sodium is bombarded with electrons accelerated through a potential difference $\mathrm{\Delta}V$, its x-ray spectrum contains emission peaks at 1.04 keV and 1.07 keV. Find the minimum value of $\mathrm{\Delta}V$ required to produce both of these peaks.

asked 2022-05-18

If a black body is radiating at ${17}^{\circ}$ C. Calculate the (a) wave length at which maximum energy will be radiated and (b) energy per unit area per second radiated by this black body.