Question

the momentum of particles associated with debroglie wave length of 1A*​

Answer 1

Answer:

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The De-Broglie Hypothesis states that matter particles also behave like waves. So, in a way, all matter particles also possess the Wave-Particle Duality.

The De-Broglie Wavelength (\lambda)(λ) of a particle is given as:

\boxed{\lambda = \frac{h}{p}}

λ=

p

h

Here, h is the Planck's Constant, and p is Momentum of the Particle.

Also, the value of Planck's Constant is :

\boxed{h = 6.626 \times 10^{-34} \, \, J \, s}

h=6.626×10

−34

Js

Also, we are given that the Wavelength is :

Now, we can find the momentum as follows:

\begin{lgathered}\lambda = \frac{h}{p} \\ \\ \\ \implies p = \frac{h}{\lambda} \\ \\ \\ \implies p = \frac{6.626 \times 10^{-34}}{10^{-10}} \\ \\ \\ \implies \boxed{p = 6.626 \times 10^{-24} \, \, kg \, m / s}\end{lgathered}

λ=

p

h

⟹p=

λ

h

⟹p=

10

−10

6.626×10

−34

⟹

p=6.626×10

−24

kgm/s

Thus, the Momentum of the particle is 6.626 \times 10^{-24} \, \, kg \, m/s6.626×10

−24

kgm/s

Hope it helps^_^