Question

Calculate the de Broglie wavelength associated with a helium atom in a helium gas sample at $30K$ and $1atm$ pressure.(Take $h=6\times {10}^{-34}$, $N_A=6\times {10}^{23}$ and Boltzman constant $1.5\times {10}^{-23}$ in their standard unit)

• A. $7.3\times {10}^{-11}$ metre.
• B. $2.0\times {10}^{-10}$ metre.
• C. $7.3\times {10}^{-10}$ metre.
• D. $3.6\times {10}^{-11}$ metre.

Answer 1

The correct option is B $2.0\times {10}^{-10}$ 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, $m=\frac{4\times {10}^{-3}}{6\times {10}^{23}}=6.66\times {10}^{-27}kg$
$T=30K$ and $K=1.5\times {10}^{-23}J{K}^{-1}$
$\therefore \lambda =\frac{h}{\sqrt{3mKT}}=\frac{6\times {10}^{-34}}{\sqrt{3\times 6.66\times {10}^{-27}\times 1.5\times {10}^{-23}\times 30}}=2.0\times {10}^{-10}m$