Question

Calculate the de-Broglie wavelength of an electron moving with one fifth of the speed of light. Neglect relativistic effects. ($h=6.63\times {10}^{-34}J.s.,c=3\times {10}^{8}m/s$, mass of electron $=9\times {10}^{-31}kg$)

Answer 1

By de-Broglie's theory, the wavelength of any particle $\lambda =\frac{h}{p}$ and $p=m_V$ $\therefore \lambda =\frac{h}{m_V}$

${}_V=\frac{c}{5}=\frac{3\times {10}^8}{5}$

$\therefore \lambda =\frac{6.63\times {10}^{-34}\times 5}{9\times {10}^{-31}\times 3\times {10}^8}=1.23\times {10}^{-11}m$

Answer:- $1.23\times {10}^{-11}metres$