Question

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

Answer 1

Step 1: Given Data

The velocity of the helium atoms $v=2\times {10}^{4}m/s$

Let the Planck's constant be $h=6.626\times {10}^{-34}kg{m}^2/s$

Let the total mass of helium be $m$.

Step 2: Calculate the total mass

The gram atomic mass of helium $=4g/mol$

$1mol=6.022\times {10}^{23}$ atoms

Therefore, the mass of $6.022\times {10}^{23}$ atoms of helium $=4g=4\times {10}^{-3}kg$

Hence, the mass of one helium atom,

$m=\frac{4\times {10}^{-3}}{6.022\times {10}^{23}}$

$=6.64\times {10}^{-27}kg$

Step 3: Calculate the Wavelength

We know that DeBroglie wavelength is given as

$\lambda =\frac{h}{mv}$

Upon substituting the values we get,

$\lambda =\frac{6.626\times {10}^{-34}}{6.64\times {10}^{-27}\times 2\times {10}^4}$

$=4.98\times {10}^{-8}m$

Hence, the wavelength of particles constituting the beam is $4.98\times {10}^{-8}m$.