Answer to Question #132234 in Mechanics | Relativity for Max

Question

Find out de-broglie wavelength of proton whose kinetic energy is 10KeV.

Answer 1

solution

given data

mass of proton (m)$=1.67\times10^{-27}kg$

charge of proton (q)=$1.6\times10^{-19}C$

1eV=$1.6\times10^{-19}V$

kinetic energy of proton (KE)=10KeV

kinetic energy is given by

$KE=\frac{p^2}{2m}$ ........eq.1

de-broglie wavelength is given by

$\lambda=\frac{h}{p}$ ........eq.2

by using equation 1 and 2 we get

$\lambda=\frac{h}{\sqrt{2mKE}}$

by putting the value of h ,m and KE

$\lambda=\frac{6.625\times10^{-34}}{\sqrt{2\times1.67\times10^{-27}\times10^4\times1.6\times10^{-19}}}$

$\lambda=\frac{6.625\times10^{-13}}{\sqrt{2}\times1.6}$

$\lambda=2.93\times10^{-13}m$

this is comparable of size relative order to size of nucleus.