Question

Calculate the de Broglie wavelength associated with a helium atom in a helium gas sample at ${27}^{o}C$ and $1atm$ pressure.

• A. $7.3\times {10}^{-11}$ metre.
• B. $3.6\times {10}^{-10}$ metre.
• C. $7.3\times {10}^{-10}$ metre.
• D. $3.6\times {10}^{-11}$ metre.

Answer 1

The correct option is A $7.3\times {10}^{-11}$ metre.

Energy $E=\frac{3}{2}KT$ where $K=$ Boltzmann constant.

$\frac{3}{2}KT=\frac{1}{2}m{v}^2⟹mv=\sqrt{3mKT}$

$\therefore \lambda =\frac{h}{mv}=\frac{h}{\sqrt{3mKT}}$

For Helium atom, $m=\frac{4\times {10}^{-3}}{6.02\times {10}^{23}}=6.64\times {10}^{-27}kg$

$T=27+273=300K$ and $K=1.38\times {10}^{-23}J{K}^{-1}$

$\therefore \lambda =\frac{6h}{\sqrt{3mKT}}=\frac{6.6\times {10}^{-34}}{\sqrt{3\times 6.64\times {10}^{-27}\times 1.38\times {10}^{-23}\times 300}}\approx 7.3\times {10}^{-11}m$