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$\text{Hint: Wavelength is inversely proportional to mass}$\n\n$\text{The de Broglie wavelength is given by:}$\n\n$\lambda = \frac{h}{p}$\n\n$\text{where } h \text{ is Planck's constant and } p \text{ is the momentum.}$\n\n$\text{Since } p = mv \text{, where } m \text{ is the mass and } v \text{ is the velocity:}$\n\n$\lambda = \frac{h}{mv}$\n\n$\text{Given that all particles have equal velocity, we can see that:}$\n\n$\lambda \propto \frac{1}{m}$\n\n$\text{This means the de Broglie wavelength is inversely proportional to the mass of the particle. The particle with the smallest mass will have the largest wavelength.}$\n\n$\text{Let's compare the masses:}$\n\n$\text{The mass of CO}_2 \text{ molecule is 44 amu.}$\n\n$\text{The mass of NH}_3 \text{ molecule is 17 amu.}$\n\n$\text{The mass of electron is } 9.10938356 \times 10^{-31} \text{ kg}$\n\n$\text{The mass of proton is } 1.6726219 \times 10^{-27} \text{ kg}$\n\n$\text{Converting all to the same units for comparison:}$\n\n$\text{1 amu } \approx 1.66054 \times 10^{-27} \text{ kg}$\n\n$\text{CO}_2\text{: } 44 \text{ amu } \approx 7.30638 \times 10^{-26} \text{ kg}$\n\n$\text{NH}_3\text{: } 17 \text{ amu } \approx 2.82292 \times 10^{-26} \text{ kg}$\n\n$\text{Electron: } 9.10938356 \times 10^{-31} \text{ kg}$\n\n$\text{Proton: } 1.6726219 \times 10^{-27} \text{ kg}$\n\n$\text{Clearly, the electron has the smallest mass by several orders of magnitude. Therefore, the electron will have the largest de Broglie wavelength.}$