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:

The correct option/answer is 743 nm.

Logical and Physical Principles:
When a gas molecule or atom absorbs a photon, it gains an amount of energy equivalent to the energy of that photon. When it subsequently emits energy in the form of two photons, the total energy of the emitted photons must equal the energy of the absorbed photon, in accordance with the law of conservation of energy.
Therefore, we can write:

$$E_{\text{absorbed}}=E_1+E_2$$

where $E_{\text{absorbed}}$ is the energy of the absorbed photon, and $E_1$ and $E_2$ are the energies of the two emitted photons.

Relating Energy to Wavelength:
The energy of a photon is related to its wavelength by the Planck-Einstein relation:

$$E=\frac{hc}{\lambda }$$

where $h$ is Planck's constant, $c$ is the speed of light, and $\lambda$ is the wavelength.

Substituting this relation into our energy conservation equation, we get:

$$\frac{hc}{{\lambda }_{\text{absorbed}}}=\frac{hc}{{\lambda }_1}+\frac{hc}{{\lambda }_2}$$

By dividing both sides of the equation by $hc$, the constant terms cancel out, leaving a simple relationship between the wavelengths:

$$\frac{1}{{\lambda }_{\text{absorbed}}}=\frac{1}{{\lambda }_1}+\frac{1}{{\lambda }_2}$$

Step-by-Step Calculation:
We are given the following values:
• ${\lambda }_{\text{absorbed}}=355\text{nm}$
• ${\lambda }_1=680\text{nm}$

We need to find the value of ${\lambda }_2$:
Rearranging the equation to solve for $\frac{1}{{\lambda }_2}$:

$$\frac{1}{{\lambda }_2}=\frac{1}{{\lambda }_{\text{absorbed}}}-\frac{1}{{\lambda }_1}$$

Substitute the given values into the formula:

$$\frac{1}{{\lambda }_2}=\frac{1}{355}-\frac{1}{680}$$

Calculate the values of the fractions:

$$\frac{1}{355}\approx 0.002817{\text{nm}}^{-1}$$

$$\frac{1}{680}\approx 0.001471{\text{nm}}^{-1}$$

Subtract the two values:

$$\frac{1}{{\lambda }_2}\approx 0.002817-0.001471=0.001346{\text{nm}}^{-1}$$

Now, take the reciprocal to find ${\lambda }_2$:

$${\lambda }_2=\frac{1}{0.001346}\approx 742.94\text{nm}$$

Rounding to the nearest whole number, we get 743 nm.