Question

calculate the de Broglie wavelength of an electron that has been accelerated from rest through a potential difference of 1kv​

Answer 1

we know that ,

lambda is equal to Planck's constant (h) divided by under root 2×9.1×10^–31(mass of electrons)× 1.6× 10^-19 (value of Q)× 1000v

(1kv= 1000v)

after solving we get:-

6.6×10^-14 /17.03

= 3.87×10-11