Question

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

Answer 1

Answer: $0.498\times {10}^{-7}m$is the wavelength of the beam.

Explanation:

According to de Broglie's equation,

$\lambda =\frac{h}{m\times v}$

where,

$\lambda =wavelength$

$h=plan{k}^{\prime }sconstant=6.626\times {10}^{-34}kg{m}^2/s$

$m=mass$

Mass of helium atom $=4amu.$

Convert amu into kg as follows.

$4amu\times \frac{1g}{6.022\times {10}^{23}amu}$

$=0.664\times {10}^{-23}g$

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

Putting the values in the above equation as follows.

$\lambda =\frac{h}{m\times v}$

$=\frac{6.626\times {10}^{-34}kg{m}^2/s}{6.64\times {10}^{-27}kg\times 2\times {10}^{4}m/s}$

$=0.498\times {10}^{-7}m$

Thus, $0.498\times {10}^{-7}m$is the wavelength of the beam.