Question

Calculate the de broglie wavelength of an electron that has been accelerated from rest through potential difference of 1 kv

Answer 1

De broglie wavelength for an electron is given by:

Lambda = 12.27 angstrom/ sqr root(Potential)

Lambda = 12.27/ sqr root(1000)

= 0.38 angstrom

= 3.8 nanometres

Answer 2

when the electron has been accelerated through a potential difference of 1KV the kinetic energy it develops is1000V X e=1.602 X 10-19+3=1.602 X 10-16J.

YOU MAY NOW USE THE RELATION p2/2m=K.E i.e

p2=2 X 9.1 X 10-31 X1.602 X 10 -16=29.15 X 10-47

 p=17.07 X 10-24 kg m s-1

lamda=h/p=6.626 X 10-34/17.07 X 10-24=0.388 X 10-10 m=3.8 X 10-11 m

ALTER:

YOU MAY ALSO USE THE RELATION :

LAMBDA=1.226x10-9/(V)1/2=1.226 X 10-9/(1000)1/2=3.8X10-11m

WITH REGRDS

We know dat lamda is equal to pranks constant (h) divided by under root 2×9.1×10^-31(mass of electron)×1.6×10^-19(value of q ) ×1000 volts (1kv =1000v) On sloving we get 6.6×10^-14÷17.03 = 3.87×10^-11