A gas absorbs a photon of $$355 \, \text{nm}$$ and emits at two wavelengths. If one of the emissions is at $$680 \, \text{nm}$$, the other is at:

When a gas atom or molecule first absorbs a photon and later de-excites in two separate steps, the total energy of the two emitted photons must equal the energy of the single absorbed photon.

Photon energy is given by $$E = \frac{hc}{\lambda}$$, so energy conservation gives
$$\frac{hc}{\lambda_{\text{abs}}}= \frac{hc}{\lambda_1}+ \frac{hc}{\lambda_2}$$

The constants $$h$$ and $$c$$ cancel out, leaving
$$\frac{1}{\lambda_{\text{abs}}}= \frac{1}{\lambda_1}+ \frac{1}{\lambda_2} \quad -(1)$$

Data from the question:
$$\lambda_{\text{abs}} = 355 \,\text{nm}, \quad \lambda_1 = 680 \,\text{nm}, \quad \lambda_2 = ?$$

Substitute in equation $$-(1)$$:
$$\frac{1}{355} = \frac{1}{680} + \frac{1}{\lambda_2}$$

Solve for $$\lambda_2$$:
$$\frac{1}{\lambda_2} = \frac{1}{355} - \frac{1}{680}$$
$$\frac{1}{\lambda_2} = 0.0028169\;\text{nm}^{-1} - 0.0014706\;\text{nm}^{-1}$$
$$\frac{1}{\lambda_2} = 0.0013463\;\text{nm}^{-1}$$

Taking the reciprocal:
$$\lambda_2 = \frac{1}{0.0013463}\;\text{nm} \approx 742.8\;\text{nm}$$

Rounded to the nearest nanometre, $$\lambda_2 \approx 743 \,\text{nm}$$.

Therefore, the other emitted wavelength is $$743 \,\text{nm}$$.

Option B which is: 743 nm