A gas absorbs a photon of 355 nm and emits at two wavelengths. If one of the emissions is at 680 nm, the other is at : (1) 325 nm (2) 743 nm (3) 518 nm (4) 1035 nm

To determine the other wavelength of emission from the gas that absorbs a photon of 355 nm and emits one photon at 680 nm, we can use the principle of conservation of energy. When a gas absorbs energy in the form of a photon, it can later release that energy in different ways, including emitting photons at various wavelengths. The key here is to understand the relationship between the absorbed and emitted wavelengths.

Understanding Photon Energy

The energy of a photon is inversely related to its wavelength. This relationship is 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 the Energy of the Absorbed Photon

First, we need to calculate the energy of the absorbed photon at 355 nm:

Convert 355 nm to meters: 355 nm = 355 x 10^-9 m.

Now, substituting into the energy equation:

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

This calculation gives us the energy of the absorbed photon.

Calculating the Energy of the Emitted Photon

Next, we know that one of the emitted wavelengths is 680 nm. We can calculate its energy in a similar manner:

Convert 680 nm to meters: 680 nm = 680 x 10^-9 m.

Using the energy equation again:

E_emitted1 = (6.626 x 10^-34 J·s) * (3.00 x 10⁸ m/s) / (680 x 10^-9 m)

Finding the Other Emission Wavelength

Since the gas absorbs energy and emits it in two different wavelengths, the total energy absorbed must equal the sum of the energies of the emitted photons:

E_absorbed = E_emitted1 + E_emitted2

We can rearrange this to find the energy of the second emitted photon:

E_emitted2 = E_absorbed - E_emitted1

Calculating the Wavelength of the Second Emission

Once we have the energy of the second emitted photon, we can find its wavelength using the energy equation:

λ_emitted2 = hc/E_emitted2

Evaluating the Options

Now, let’s evaluate the given options:

• (1) 325 nm
• (2) 743 nm
• (3) 518 nm
• (4) 1035 nm

After calculating the energies and wavelengths, we find that the second emission wavelength corresponds to one of the options provided. The calculations will show that the correct answer is:

Option (2) 743 nm

In summary, by applying the principles of energy conservation and the relationship between wavelength and photon energy, we can determine the other emission wavelength from the gas after it absorbs a photon of 355 nm and emits one at 680 nm.