Question

Find the typical de Broglie wavelength associated with a $He$ atom in helium gas at room temperature (${27}^{o}C$) and $1atm$ pressure, and compare it with the mean separation between two atoms under these conditions.

• A. $r_0=6.4\times {10}^{-9}m$
• B. $r_0=3.4\times {10}^{-9}m$
• C. $r_0=4.4\times {10}^{-9}m$
• D. $r_0=5.4\times {10}^{-9}m$

Answer 1

The correct option is B $r_0=3.4\times {10}^{-9}m$

Mass of $He$ atoms,

$m=\frac{AtomicmassofHe}{Avogadr{o}^{\prime }snumber}$

$\begin{array}{c}=\frac{4}{6\times {10}^{23}}g\\ \Rightarrow m=6.67\times {10}^{-27}kg\end{array}$

De-broblie wavelength,

$\lambda =\frac{h}{p}=\frac{h}{\sqrt{3mkT}}m$

$=\frac{6.63\times {10}^{-34}}{\sqrt{3\times 6.67\times {10}^{-27}\times 1.38\times {10}^{-23}\times 300}}m$

$\lambda =7.3\times {10}^{-11}m$

Now using the kinetic gas equation for one mole of a gas

$\begin{array}{c}PV=RT=kNT\\ \Rightarrow \frac{V}{N}=\frac{kT}{P}\end{array}$

Mean Separation,

$\begin{array}{c}{r}_0={\left(\frac{Molarvolume}{Avogard{o}^{\prime }snumber}\right)}^{\frac{1}{3}}\\ ={\left(\frac{V}{N}\right)}^{\frac{1}{3}}\\ ={\left(\frac{kT}{P}\right)}^{\frac{1}{3}}\\ \Rightarrow r_0={\left(\frac{1.38\times {10}^{-23}\times 300}{1.01\times {10}^5}\right)}^{\frac{1}{3}}m\\ r_0=3.4\times {10}^{-9}m\end{array}$

So, from the above calculation we say that the mean separation between two atoms is much larger than the de-Broglie wave length.