Question

The de -Broglie wave length associated with particle of mass 10^-6 kg moving with a velocity of 10m/s is:

Answer 1

de-broglie wavelength = h / (mv)
= 6.626 × 10^-34 / (10^-6 × 10)
= 6.626 × 10^-29 m

de -Broglie wave length associated with that particle is 6.626 × 10^-29 m

Answer 2

Answer: The de-Broglie wavelength will be $6.626\times 10^{-29}m$

Explanation:

Equation used to calculate de-Broglie's wavelength is given by:

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

where,

$\lambda$ = wavelength of the particle

h = Planck's constant = $6.626\times 10^{-34}Js=6.626\times 10^{-34}kg.m^2.s^{-1}$

m = Mass of the particle = $10^{-6}kg$

v = velocity of the particle = 10 m/s

Putting values in above equation, we get:

$\lambda=\frac{6.626\times 10^{-34}kg.m^2s^{-1}}{10^{-6}kg\times 10m/s}\\\\\lambda=6.626\times 10^{-29}m$

Hence, the de-Broglie wavelength will be $6.626\times 10^{-29}m$