Question

The kinetic energy of an electron is 5 eV. Calculate the de broglie wavelength associated with it. (h = 6.6 × 10-34 Js, me = 9.1 × 10-31 kg)
(a) 5.47 Å
(b) 10.9 Å
(c) 2.7 Å
(d) None of these

Answer 1

Solution :

Given :-

▪ Kinetic energy of electron = 5eV

▪ Mass of electron = 9.1×10$^{-31}$kg

▪ Plank constant (h) = 6.6×10$^{-34}$Js

To Find :-

▪ De-broglie wavelength associated with electron.

Formula :-

▪ Formula of De-broglie wavelength associated with moving particle is given by

$\underline{\boxed{\bf{\pink{\lambda=\dfrac{h}{\sqrt{2m(KE)}}}}}}$

Calculation :-

$\implies\sf\:\lambda=\dfrac{6.6\times 10^{-34}}{\sqrt{2\times 9.1\times 10^{-31}\times 5\times 1.6\times 10^{-19}}}\\ \\ \implies\sf\:\lambda=\dfrac{6.6\times 10^{-34}}{\sqrt{145.6\times 10^{-50}}}\\ \\ \implies\sf\:\lambda=\dfrac{6.6\times 10^{-34}}{12.066\times 10^{-25}}\\ \\ \implies\sf\:\lambda=5.47\times 10^{-10}\\ \\ \implies\boxed{\bf{\purple{\lambda=5.47\:A\degree}}}$