Question

A beam of helium atoms moves with a velocity of 2*10^3m/s find wavelenght of the particle

Answer 1

mass of single He atom= 4/6.022×10²³=6.64×10^-27 using de Broglie equation wavelength = h/MV so wavelength = 6.62×10^-34/6.64×10^-27× 2×10^3= 0.5 ×10^-11(approx.)

Answer 2

Answer:

The wavelength of the particle is $4.99\times10^{-11}\ m$

Explanation:

Given that,

Velocity of helium$v =2\times10^{3}\ m/s$

Using de-broglie equation.

$\lambda=\dfrac{h}{p}$

Where, h = Planck constant

p = momentum

The momentum is the product of the mass and velocity of the particle.

{tex]\lambda= \dfrac{h}{mv}[/tex]....(I)

We know that,

One mole of He contains $6.022\times10^{23}$ atoms.

Then the mass of single He atom

$m=\dfrac{4}{6.022\times10^{23}}$

$m =6.64\times10^{-24}$

Now , put the value of m,v and h in equation (I)

$\lambda=\dfrac{6.026\times10^{-34}}{6.64\times10^{-27}\times2\times10^{3}}$

$\lambda=4.99\times10^{-11}\ m$

Hence, The wavelength of the particle is $4.99\times10^{-11}\ m$