a gas absorbs photon of 355nm and emits at two wavelengths. ifone of ion is at 680nm other will be?

When a gas absorbs a photon of 355 nm, it gets excited to a higher energy state. As it returns to a lower energy state, it can emit photons at different wavelengths. In your case, one of the emitted wavelengths is at 680 nm. To find the other emitted wavelength, we can use the concept of energy levels and the relationship between wavelength and energy.

Understanding Photon Emission

Photons are emitted when an electron in an atom or molecule transitions from a higher energy level to a lower one. The energy of a photon is inversely related to its wavelength, which can be described by the equation:

E = hc/λ

• E is the energy of the photon.
• h is Planck's constant (approximately 6.626 x 10^-34 J·s).
• c is the speed of light (approximately 3.00 x 10⁸ m/s).
• λ is the wavelength of the photon in meters.

Calculating Energy of the Absorbed Photon

First, let’s calculate the energy of the absorbed photon at 355 nm:

Convert 355 nm to meters:

355 nm = 355 x 10^-9 m

Now, plug this into the energy equation:

E_absorbed = (6.626 x 10^-34 J·s) x (3.00 x 10⁸ m/s) / (355 x 10^-9 m)

Calculating this gives:

E_absorbed ≈ 5.59 x 10^-19 J

Energy of the Emitted Photon

Now, when the gas emits a photon at 680 nm, we can calculate the energy of that photon as well:

Convert 680 nm to meters:

680 nm = 680 x 10^-9 m

Using the energy equation again:

E_{emitted (680 nm)} = (6.626 x 10^-34 J·s) x (3.00 x 10⁸ m/s) / (680 x 10^-9 m)

This results in:

E_{emitted (680 nm)} ≈ 2.91 x 10^-19 J

Finding the Energy of the Second Emission

Since the gas can emit at two wavelengths, the total energy absorbed must equal the sum of the energies of the emitted photons:

E_absorbed = E_{emitted (680 nm)} + E_{emitted (second)}

Substituting the known values:

5.59 x 10^-19 J = 2.91 x 10^-19 J + E_{emitted (second)}

Solving for the energy of the second emitted photon:

E_{emitted (second)} = 5.59 x 10^-19 J - 2.91 x 10^-19 J

E_{emitted (second)} ≈ 2.68 x 10^-19 J

Calculating the Wavelength of the Second Emission

Now, we can find the wavelength corresponding to this energy using the rearranged energy equation:

λ_second = hc/E_{emitted (second)}

Substituting the values:

λ_second = (6.626 x 10^-34 J·s) x (3.00 x 10⁸ m/s) / (2.68 x 10^-19 J)

This calculation yields:

λ_second ≈ 740 nm

Summary of Results

In summary, when a gas absorbs a photon of 355 nm and emits at 680 nm, the other emitted wavelength is approximately 740 nm. This process illustrates the fascinating interplay between energy absorption and emission in atomic and molecular systems.