Question

if the plancks constant h=6.6x10'-34 js the de Broglie wavelength of a particle having momentum 3.3x10'_24kg ms'_1will be?

Answer 1

Answer: The De-Broglie wavelength of the given particle is $2\times 10^{-10}m$

Explanation:

To calculate the wavelength of a particle, we use the equation given by De-Broglie's wavelength, which is:

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

where,

$\lambda$ = De-Broglie's wavelength = ?

h = Planck's constant = $6.6\times 10^{-34}Js$

p = momentum of the particle = $3.3\times 10^{-24}kg.m/s$

Putting values in above equation, we get:

$\lambda=\frac{6.6\times 10^{-34}Js}{3.3\times 10^{-24}kgm/s}\\\\\lambda=2\times 10^{-10}m$

Hence, the De-Broglie wavelength of the given particle is $2\times 10^{-10}m$