Carbon absorbs energy at a wavelength of 150. nm.

I. What is the energy (in J) of one photon of 150. nm light?
[ Select ]
1.47 × 10^–35 J
9.94 × 10^–32 J
1.32 × 10^–18 J
9.94 × 10^–41 J
1.32 × 10^–27 J

II. If a sample of carbon emits 1.98 × 105 J of energy, how many carbon atoms are present in the sample assuming that each atom emits one photon?
[ Select ]
1.50 × 10^23 atoms
2.99 × 10^38 atoms
1.32 × 10^12 atoms
1.50 × 10^30 atoms
2.62 × 10^–13 atoms

11A

Question

Carbon absorbs energy at a wavelength of 150. nm.

I. What is the energy (in J) of one photon of 150. nm light? 
[ Select ]
1.47 × 10^–35 J
9.94 × 10^–32 J
1.32 × 10^–18 J
9.94 × 10^–41 J
1.32 × 10^–27 J

II. If a sample of carbon emits 1.98 × 105 J of energy, how many carbon atoms are present in the sample assuming that each atom emits one photon? 
[ Select ]
1.50 × 10^23 atoms
2.99 × 10^38 atoms
1.32 × 10^12 atoms
1.50 × 10^30 atoms
2.62 × 10^–13 atoms

11A

BlackTom AI · Accepted Answer

Starting with the prompt, we have a two-part problem about carbon absorbing light at a wavelength of 150. nm and then an energy-emission scenario.

Part I asks for the energy of a single photon with wavelength 150. nm. The energy per photon is given by E = h c / λ, where h is Planck's constant and c is the speed of light. Converting the wavelength: λ = 150 nm = 150 × 10^−9 m. Using h = 6.626 × 10^−34 J·s and c = 3.00 × 10^8 m/s, compute E = (6.626×10^−34)(3.00×10^8) / (150×10^−9) J. Multiplying the numerator gives about 1.9878×10^−25 J·m, and dividing by 1.50×10^−7 m yields E ≈ 1.325×10^−18 J, which is 1.32 × 10^−18 J to two significant figures.

Option-wise check for Part I:
- 1.47 × 10^−35 J: far too small by many orders of magnitude; unlikely for a photon energy at this wavelength.
- 9.94 × 10^−32 J: still far too small; off by about 14 orders of magnitude.
- 1.32 × 10^−18 J: matches the calculated value; this is consistent with E = h c / λ for 150 nm.
- 9.94 × 10^−41 J: extremely tiny; not correct for this wavelength.
- 1.32 × 10^−27 J: too large by about 9 orders of magnitude.
Thus, the correct Part I energy per photon is 1.32 × 10^−18 J.

Part II asks: If a sample of carbon emits 1.98 × 10^5 J of energy, how many carbon atoms are present, assuming each atom emits one photon? The number of photons emitted equals the total energy divided by the energy per photon (as found in Part I). So N = (1.98 × 10^5 J) / (1.32 × 10^−18 J per photon) = (1.98 / 1.32) × 10^(5 + 18) ≈ 1.50 × 10^23 photons (and thus atoms).

Option-wise check for Part II:
- 1.50 × 10^23 atoms: matches the calculation above; this is the correct count of atoms given one photon per atom.
- 2.99 × 10^38 atoms: far too large; would require a much larger total energy or a much smaller photon energy.
- 1.32 × 10^12 atoms: far too small; off by 11 orders of magnitude.
- 1.50 × 10^30 atoms: too large by about 7 orders of magnitude.
- 2.62 × 10^−13 atoms: nonsensical in this context (fractional atom count for a macroscopic sample).

Overall, Part I selects the energy per photon as 1.32 × 10^−18 J, and Part II selects 1.50 × 10^23 atoms as the number of carbon atoms emitting one photon each. The incorrect options are ruled out by mismatches with the fundamental E = h c / λ relationship for Part I and by straightforward division for Part II.

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