Question

Calculate the momentum of a particle which has a de-Broglie wavelength of 1A°.
[h = 6.626 × 10⁻³⁴ kg m2 s⁻¹]

Answer 1

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 $(\lambda)$ of a particle is given as:

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

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

Also, the value of Planck's Constant is :

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


Also, we are given that the Wavelength is :

$\lambda = 1 \, \, \AA = 10^{-10} \, \, m$

Now, we can find the momentum as follows:

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


Thus, the Momentum of the particle is $6.626 \times 10^{-24} \, \, kg \, m/s$


Hope it helps
Purva
Brainly Community

Answer 2

Answer:

the value of plank constant is h 6.64*10^-34