Question

A beam of helium atoms moves with a velocity of $2\times {10}^4{ms}^{-1}$. Find the wavelength of particles constituting with the beam.

Answer 1

According to de-Broglie relation: $\lambda$ = $\frac{h}{P}$ =$\frac{h}{mv}$

1 mol or 4 g of helium contains $6.022\times {10}^2$${}^3$ atoms, then the mass of single He atom is, $6.64\times {10}^{-}$${}^2$${}^4$ g

So, the wavelength=

$\lambda$ = $\frac{h}{mv}$ = $6.672\times {10}^{-34}$ Kg $m^2{s}^{-}$${}^1$ / $6.64\times {10}^2$${}^7$ Kg $\times 20000$ m$s^{-}$${}^1$

=$4.99\times {10}^{-}$${}^1$${}^2$ m

$\lambda$ = 4.99 pm