Question

De Broglie wavelength for an average helium atom in a furnace at 400 K is $k\times {10}^{-10}m$. Find the value of k. Here, Given the mass of helium $=4.002amu$

Answer 1

Given : $T=400K$ $m=4.002amu=4.002\times 1.67\times {10}^{-27}kg$
Velocity of He atom $v=\sqrt{\frac{3kT}{m}}$
$v=\sqrt{\frac{3\times 1.38\times {10}^{-23}\times 400}{4.002\times 1.67\times {10}^{-27}}}=1574m/s$
de-Broglie wavelength $\lambda =\frac{h}{mv}$
$⟹\lambda =\frac{6.63\times {10}^{-34}}{4.002\times 1.67\times {10}^{-27}\times 1574}=0.63\times {10}^{-10}m$