The wavelength for these sodium vapor lamps is 570 nm, and 45.8 mg of sodium atoms emitting one photon each releases about 1.9×1021.9\times 10^{2}1.9×102 J of energy, which rounds to 190.0 J.

How to get this result (briefly)

  1. Use the wavelength from the earlier question From the referenced problem set, the sodium vapor lamp in “question 13” (same setup as your “question 7”) emits light of wavelength 570 nm.
  1. Energy per photon Use E=hcλE=\dfrac{hc}{\lambda}E=λhc​ with:
    • h=6.626×10−34h=6.626\times 10^{-34}h=6.626×10−34 J·s
    • c=3.00×108c=3.00\times 10^{8}c=3.00×108 m/s
    • λ=570×10−9\lambda =570\times 10^{-9}λ=570×10−9 m

This gives an energy per photon on the order of 3.5×10−193.5\times 10^{-19}3.5×10−19 J (same method as shown for similar sodium-lamp questions).

  1. Number of sodium atoms in 45.8 mg
    • Mass of Na sample: 45.8 mg =4.58×10−2=4.58\times 10^{-2}=4.58×10−2 g.
    • Moles of Na: 4.58×10−2textg22.99textg/mol≈1.99×10−3\dfrac{4.58\times 10^{-2}\\text{g}}{22.99\\text{g/mol}}\approx 1.99\times 10^{-3}22.99textg/mol4.58×10−2textg​≈1.99×10−3 mol.
    • Number of atoms: 1.99×10−3textmol×6.022×1023textatoms/mol≈1.20×10211.99\times 10^{-3}\\text{mol}\times 6.022\times 10^{23}\\text{atoms/mol}\approx 1.20\times 10^{21}1.99×10−3textmol×6.022×1023textatoms/mol≈1.20×1021 atoms (and thus photons, since each atom emits one).
  1. Total energy Multiply energy per photon by the number of photons:

Etotal≈(3.5×10−19textJ/photon)×(1.20×1021textphotons)≈1.9×102textJE_{\text{total}}\approx (3.5\times 10^{-19}\\text{J/photon})\times (1.20\times 10^{21}\\text{photons})\approx 1.9\times 10^{2}\\text{J}Etotal​≈(3.5×10−19textJ/photon)×(1.20×1021textphotons)≈1.9×102textJ

Rounded to the nearest tenth, this is 190.0 J.

Answer (as requested):

1.9×102textJ or 190.0textJ\boxed{1.9\times 10^{2}\\text{J};\text{or};190.0\\text{J}}1.9×102textJor190.0textJ​