A gas absorbs a photon of wavelength 355 nm and then emits radiation at two longer wavelengths. If one of the emission lines is observed at 680 nm, what is the wavelength of the other emission line?

Introduction / Context:
This question examines your understanding of basic quantum theory and the relationship between photon energy and wavelength. When a gas absorbs a photon, an electron in an atom or molecule is promoted to a higher energy level. It can then return to lower energy levels in one or more steps, emitting photons of specific wavelengths. The sum of the energies of emitted photons must equal the energy of the absorbed photon. Using the relation between energy and wavelength, you can determine an unknown emission wavelength when the absorbed wavelength and one emission wavelength are known.

Given Data / Assumptions:

• The gas absorbs a photon with wavelength lambda_abs = 355 nm.
• The gas emits two photons when returning to the ground state or a lower state.
• One emitted photon has wavelength lambda_1 = 680 nm.
• The other emitted photon has wavelength lambda_2, which is unknown.
• Energy levels are assumed to be fixed and there is no loss of energy as heat.
• The energy of a photon is given by E = h * c / lambda.

Concept / Approach:
Photon energy is inversely proportional to wavelength. For a photon, E = h * c / lambda, where h is Planck constant and c is speed of light. When an atom absorbs a photon, it gains energy E_abs. When it relaxes in two steps and emits two photons, their energies are E_1 and E_2. By conservation of energy, E_abs = E_1 + E_2. Because h and c are constants, you can work conveniently with inverse wavelengths (called wave numbers). The relationship becomes 1 / lambda_abs = 1 / lambda_1 + 1 / lambda_2. Solving this equation for lambda_2 gives the missing wavelength.

Step-by-Step Solution:
Step 1: Write the conservation of energy using wave numbers: 1 / lambda_abs = 1 / lambda_1 + 1 / lambda_2. Step 2: Substitute lambda_abs = 355 nm and lambda_1 = 680 nm. Step 3: Compute 1 / lambda_2 = 1 / 355 nm minus 1 / 680 nm. Step 4: Evaluate 1 / 355 nm approximately as 0.0028169 nm^-1 and 1 / 680 nm approximately as 0.0014706 nm^-1. Step 5: Subtract to get 1 / lambda_2 approximately equal to 0.0028169 minus 0.0014706 which is about 0.0013463 nm^-1. Step 6: Take the reciprocal: lambda_2 is about 1 / 0.0013463 nm which is very close to 743 nm. Step 7: Compare the calculated value with the options and select 743 nm.

Verification / Alternative check:
As a quick check, convert the three wavelengths into relative energies. The smallest wavelength 355 nm should correspond to the highest energy. The emission wavelengths must be longer than 355 nm and their energies must add up to the absorbed energy. A 680 nm photon clearly has less energy than a 355 nm photon. The second photon must therefore also have energy smaller than that of the absorbed photon but large enough so that E_1 + E_2 equals E_abs. A wavelength of 743 nm is longer than 680 nm, so its energy is slightly lower. If you compute E_355 minus E_680, the difference should be nearly equal to E_743. This confirms that 743 nm is consistent with energy conservation, while the other options are not.

Why Other Options Are Wrong:
An emission wavelength of 518 nm would have energy too high when combined with the energy of a 680 nm photon, exceeding the absorbed energy at 355 nm. A wavelength of 325 nm would have even more energy than the absorbed photon and cannot be an emission wavelength in this process. A wavelength of 1035 nm would have a very low energy; combined with the 680 nm photon it would not account for the full absorbed energy, leaving an energy gap. Hence these options violate energy conservation in the transition scheme.

Common Pitfalls:
Common mistakes include adding wavelengths directly instead of working with energies or inverse wavelengths. Some students forget that energy is proportional to 1 / lambda, not to lambda itself. Others mistakenly use 1 / lambda_2 = 1 / lambda_1 minus 1 / lambda_abs, reversing the relation. Rounding too aggressively in intermediate steps can also lead to slightly incorrect values that do not match any option. To avoid these issues, always start from energy conservation, convert it to an equation in inverse wavelengths and maintain enough significant figures until the final step.

Final Answer:
The wavelength of the second emitted photon is 743 nm.