Question

Calculate the De Broglie wavelength of an electron with an energy of 487eV

Give your answers in nm to 3 decimal place

Answer 1

(amk De-Broglie wavelength (1) h m= mass of electron = 9.1 x 10-31 Kg K= Kinetic energy = 487 ev n = planer's constant 6.626 X10-34 So, da 6-626 *10-34 J.S 2x 9.1X10-31 K8 X 487 ev J.S CS Scanned with CamScanner

6.626 X10-34 J.S so, A= 2x 9-1X10-31 kg x 487% 1.6X10-19 J ::de 6.626 X 10-34 m H1008 x 10-23 id=5.543 X 10-11 m :1-0.055 X10-9m 10.055 nm -angwer unit conversion, J. S J. S V K8.J 13=K8m25-27 KO. K8. mag-2 J.S 11 Kg m st J K 8 ms w.m x [13= N.m am Inm = 10-9m CS Scanned with CamScanner