Firework Color Emissions: Determining Energy of Strontium and Barium Salts

[VenomHowell15]

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?

 

[lightarrow]

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".