Question

What will be the linear momentum p, mass m, and kinetic energy EK of a particle with velocity of 1250 m s^-1 and wavelength of
828 x 10^-24 nm?

Does the behavior of this particle correspond to a macroscopic or
microscopic system? Justify your answer.

Answer 1

Answer:

The de broglie relation says

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

putting in $v = 1250ms^{-2}$, $\lambda = 828*10^{-24}nm$, and $h =6.63*10^{-34}$ we get:

$m = \dfrac{6.63*10^{-34}}{(828*10^{-24}*10^{-9}m) 1250ms^{-2}}$

$\boxed{m=6.41*10^{-7}kg = 64\mu g}$

The kinetic energy of the particle is

$\dfrac{1}{2}mv^2= \dfrac{1}{2}(6.41*10^{-7})(1250)^2$

$= 0.5007J$

And the  linear momentum of the particle is

$mv = 6.41*10^{-7}(1250)\\\\= 8.01*10^{-4}kgms^{-2}$

The wavelength of the particle is very small, in fact it is $\approx 10^{15}$ times smaller than the diameter of the proton; therefore, for all intents and purposes, the behavior of the particle is macroscopic.