1. Use conservation of energy

When one photon of wavelength $355\,\text{nm}$ is absorbed, its energy is

$E_{\text{abs}}=\frac{hc}{355}$

If the gas emits two photons, one at $680\,\text{nm}$ and the other at $\lambda$, then

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

So,

$\frac{hc}{355}=\frac{hc}{680}+\frac{hc}{\lambda}$

Cancel $hc$:

$\frac{1}{355}=\frac{1}{680}+\frac{1}{\lambda}$

2. Solve for $\lambda$

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

Take LCM in compact form:

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

$\frac{1}{\lambda}=\frac{325}{241400}$

Hence,

$\lambda=\frac{241400}{325}$

$\lambda\approx 742.8\,\text{nm}$

So the other emission wavelength is approximately

$\boxed{743\,\text{nm}}$

3. Match with options

• A: $325\,\text{nm}$
• B: $743\,\text{nm}$
• C: $518\,\text{nm}$
• D: $1035\,\text{nm}$

Therefore, the correct option is:

$\boxed{\text{B}}$