Firework Color Emissions: Determining Energy of Strontium and Barium Salts

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In summary, the reaction between a strong oxidizing agent and an organic compound in a fireworks device causes certain salts, such as strontium and barium salts, to emit specific colors. Strontium salts emit an intense red color at 631 nm, while barium salts emit a blue-green color at 493 nm. To calculate the energy of these emissions, the equation E=hv is used, assuming that each atom emits one photon. For 1.0 g of SrCl2, approximately 119.8 J of photons would be emitted, and a similar calculation can be done for BaCl2.
  • #1
VenomHowell15
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1.) In a typical fireworks device, the hat of the reaction between the strong oxidizing agent, such as KCLO4, and an organic compound excites certain salts, which emit specific colours. Strontium salts have an intense emission at 631 nm, and barium salts have one at 493 nm.

Part a was simple enough... Simply what are the colours. Strontium salts are essentially orange, and barium blue-green. Part b is as follows:

What is the energy of these emissions for 1.0g each of the chloride salts of Sr and Ba? Assume that all heat released is converted to light emitted.

I've been scanning through my lecture notes looking for an applicable equation, but E=mc^2 is reserved for nuclear reactions or relativistic rest energy, and I doubt that's what I'm looking for here... But I somewhat doubt doing a heat of formation thing would work here either. Any pointers?
 
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  • #2
VenomHowell15 said:
1.) In a typical fireworks device, the hat of the reaction between the strong oxidizing agent, such as KCLO4, and an organic compound excites certain salts, which emit specific colours. Strontium salts have an intense emission at 631 nm, and barium salts have one at 493 nm.

Part a was simple enough... Simply what are the colours. Strontium salts are essentially orange, and barium blue-green. Part b is as follows:

What is the energy of these emissions for 1.0g each of the chloride salts of Sr and Ba? Assume that all heat released is converted to light emitted.

I've been scanning through my lecture notes looking for an applicable equation, but E=mc^2 is reserved for nuclear reactions or relativistic rest energy, and I doubt that's what I'm looking for here... But I somewhat doubt doing a heat of formation thing would work here either. Any pointers?

1.0 g of Sr(Cl)2 = 1/(87.6+70.9) = 6.3*10^(-3) mol of Sr

1.0 g of Ba(Cl)2 = 1/(137.3+70.9) = 4.8*10^(-3) mol of Ba

631 nm --> v = frequency = 3*10^8/6.31*10^(-7) = 4.8*10^14 Hz

493 nm --> v = 3*10^8/4.93*10^(-7) = 6.1*10^14 Hz

One photon emitted: E = hv

If every atom emits exactly one photon (I assume the question was meant in this way, but I would control this), then 6.3*10^(-3) mol of Sr atoms = [6.3*10^(-3)]*[6.0*10^23] = 3.78*10^21 Sr atoms would emit:

[3.78*10^21]*hv = [3.78*10^21]*6.6*10^(-34)*4.8*10^14

= 119.8 J of photons

You can do the same with Ba.

Make me know if the question was meant in the way I have assumed.
P.S. I would call the Sr emission "red" not "orange".
 
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  • #3


To determine the energy of the emissions for 1.0g each of the chloride salts of Sr and Ba, we can use the equation E = hc/λ, where E is the energy, h is Planck's constant (6.626 x 10^-34 J*s), c is the speed of light (3.00 x 10^8 m/s), and λ is the wavelength of the emission.

For strontium chloride, the emission occurs at 631 nm, so we can plug that into the equation to get:

E = (6.626 x 10^-34 J*s)(3.00 x 10^8 m/s)/(631 x 10^-9 m) = 3.14 x 10^-19 J

This is the energy of the emission for 1.0g of strontium chloride.

Similarly, for barium chloride, the emission occurs at 493 nm, so we can plug that into the equation to get:

E = (6.626 x 10^-34 J*s)(3.00 x 10^8 m/s)/(493 x 10^-9 m) = 4.03 x 10^-19 J

This is the energy of the emission for 1.0g of barium chloride.

It is important to note that this calculation assumes that all of the heat released during the reaction is converted to light emitted. In reality, some of the energy may be lost as heat or sound, so these values may be slightly higher than what is actually observed. Nevertheless, this equation can give us a good estimate of the energy of the emissions for these salts.
 

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