Question

Calculate the de broglie wavelength of an electron moving at 1 speed of light

Answer 1

According to be Brogile equation λ=hmv

Mass of electron =9.1×10−31kg

Planck's constant =6.626×10−34kgm2s−1

Velocity of electron =1 % of speed of light

=3.0×108×0.01=3×106ms−1

wave length of electron (λ)=hm

=(6.626×10−34kgm2s−1)(9.1×10−31kg)×(3×106ms−1)

=2.43×10−10m

Answer 2

Given that,

Mass of electron, $m=9.1\times 10^{-31}\ kg$

Speed of electron, $v=c=3\times 10^8\ m/s$

To find,

De Broglie wavelength of an electron.

Solution,

The De-Broglie wavelength of an object is given by :

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

h is Plank's constant

$\lambda=\dfrac{6.63\times 10^{-34}}{9.1\times 10^{-31}\times 3\times 10^8}\\\\\lambda=2.42\times 10^{-12}\ m$

So, the De-Broglie wavelength of the bullet is $2.42\times 10^{-12}\ m$.

Learn more,

De-Broglie wavelength