Dual Nature of Radiation and Matter News

11.16 Wavelength of electron, ${\lambda }_e$ = Wavelength of proton, ${\lambda }_p$ = 1.0 nm = 1 $*{10}^{-9}$ m

Planck's constant, h = 6.626 $*{10}^{-34}$ Js

From De Broglie wavelength relation, $\lambda =\frac{h}{p},$ where p = momentum

$p$ = $\frac{h}{\lambda }$ = $\frac{6.626\mathrm{}*{10}^{-34}}{1\mathrm{}*{10}^{-9}}$ = 6.626 $*{10}^{-25}$ kg.m/s. Since ${\lambda }_e={\lambda }_p$ , their momentum will be also equal.

The energy of photon is given by the relation: $E$ = $\frac{hc}{\lambda },$

where c = speed of light = 3 $*{10}^8$ m/s

$E$ = $\frac{6.626\mathrm{}*{10}^{-34}*3*{10}^8}{1\mathrm{}*{10}^{-9}}$ Js = $\frac{6.626\mathrm{}*{10}^{-34}*3*{10}^8}{1\mathrm{}*{10}^{-9}*1.6*{10}^{-19}}$ eV = 1242.38 eV = 1.242 keV

Kinetic energy of electron, having momentum p is given by the relation

$E_k$ = $\frac{1}{2}\frac{p^2}{m}$ where m = mass of electron = 9.1 $*{10}^{-31}$ kg.

Hence, $E_k$ = $\frac{1}{2}\frac{{(6.626\mathrm{}*{10}^{-25})}^2}{9.1\mathrm{}*{10}^{-31}}$ J = 2.412 $*{10}^{-19}$ J = $\frac{2.412*{10}^{-19}}{1.6*{10}^{-19}}$ eV = 1.51 eV