Question

A beam of helium atoms moves with a velocity of 2* 10^4m/s. find the wavelength of particles constituting with the beam

Answer 1

As velocity of the particle is 2 * 10^4 m/s ,
Wavelength = h/ mv
= 6.626 * 10^-34/9.1 *
10-31 * 2 *10^4
= 0.3 * 10^ -7
HOPE IT HELPS !!!

Answer 2

Answer:

The wavelength of particles constituting with the beam is $4.98\times 10^{-12} m$.

Explanation:

De-Broglie wavelength is calculated by using the formula:

$\Lambda=\frac{h}{mv}$  

where,

$\lambda$ = wavelength of the particle

h = Planck's constant = $6.626\times 10^{-34}Js$

m = mass of helium atom :

$\frac{4 g/mol}{6.022\times 10^{23} mol^{-1}}=6.647\times 10^{-27} kg$

v = velocity of particle=$2\times 10^4 m/s$

$\lambda =\frac{6.626\times 10^{-34}Js}{6.647\times 10^{-27} kg\times 2\times 10^4 m/s}=4.98\times 10^{-12} m$

The wavelength of particles constituting with the beam is $4.98\times 10^{-12} m$.