# An atom moving at its root-mean-square speed at 20 °C has a wavelength of 32.8 pm. Identify the atom

An atom moving at its root-mean-square speed at $$\pu{20 °C}$$ has a wavelength of $$\pu{3.28E-11 m}$$. Identify the atom.

Here's what I did: $$u_\mathrm{rms} = \sqrt{\frac{3RT}{M}}$$ and $$\lambda = \frac{h}{mu_\mathrm{rms}} \implies u_\mathrm{rms} = \frac{h}{m\lambda} \implies \sqrt{\frac{3RT}{M}}=\frac{h}{m\lambda},$$ where $$M$$ and $$m$$ are the masses of $$1$$ mole gas (in $$\pu{kg}$$) and $$1$$ atom of gas (in $$\pu{kg}$$), respectively. Here nothing is written about the gas, so I assume the molecule to be monoatomic only. Let the molar mass of the gas be $$M$$ (in $$\pu{kg mol-1}$$). So the mass of $$1$$ molecule is $$M/N_\mathrm{A}$$ (in $$\pu{kg}$$). \begin{align} \sqrt{\frac{3RT}{M}} &= \frac{N_\mathrm{A}h}{M\lambda}\\ \implies \sqrt{\frac{3×0.0821×293.15}{M}} &= \frac{6.022×10^{23}×6.63×10^{-34}}{M×3.28×10^{-11}}\\ \implies \sqrt{\frac{72.2}{M}} &= \frac{4×10^{-10}}{M×3.28×10^{-11}}\\ \implies \sqrt{\frac{72.2}{M}} &= \frac{12.17}{M}\\ \implies \frac{72.2}{M} &= \frac{148.17}{M^2}\\ \implies {72.2} &= \frac{148.17}{M}\\ \implies M &= \frac{148.17}{72.2} \approx 2 \end{align}

So $$M=2$$ and the molar mass of the gas is $$\pu{2 kg mol-1}$$. But there is no gas whose molecular mass is $$\pu{2 kg mol-1}$$.

So where am I wrong? Any other method is appreciated.

• Why G? We not to keep M? // You have fast-forwarded to the result too fast. To say what you did wrong, it must be seen what you did. Oct 19 at 3:40
• I have not obtained 2. Check for numerical errors. Oct 19 at 3:55
• Why not to use $R=\pu{8.314 JK-1mol-1}$? Oct 19 at 4:18
• Oh yes, Thanks @Poutnik I messed up with the units Oct 19 at 4:37
• Welcome to the site. Note, chemical information may be advantageously formatted using on ChemSE with mhchem. Take moment to familiarize with this. You are encouraged to use it in the body of questions, answers, and comments. Because it is something special not all web browsers understand well, do not use it in the title of questions or answers. And 2) if you carry units, you may check the results for plausibility by dimensional analysis, too. Oct 19 at 12:28

The algebra/formulas looks correct, a tip would be to arrange $$\sqrt{\frac{3RT}{M}}=\frac{N_\mathrm{A}h}{M\lambda}$$ to $$M=\frac{N_\mathrm{A}^2h^2}{3RT\lambda^2},$$ then swapping in the numbers, it helps to avoid any errors propagating through if you miss write a number.
If you use the second equation for $$M$$, with the constants \begin{align} R &= \pu{8.314 J K-1 mol-1}\\ h &= \pu{6.626E-34 J s}\\ N_\mathrm{A} &= \pu{6.02E23 mol-1}\\ T &= \pu{293 K}\\ \lambda &= \pu{3.28E-11 m} \end{align} you get $$M=\pu{2.02E-2 kg mol-1},$$ which is $$\pu{20 g mol-1}$$ which is the mass of neon.
A side note: if the mass was $$\pu{2 g mol-1}$$ then the gas could have been hydrogen as you don't need to assume it is monoatomic.
• Try $\pu{8.314 JK-1mol-1}$ to get $\pu{8.314 JK-1mol-1}$ // $h=\pu{6.626E-34 Js}$ as $h=\pu{6.626E-34 Js}$ Oct 19 at 5:54