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

Question

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.

Answer 1

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.