Question

The wavelength of electrons accelerated from rest through a potential difference of $40kV$ is $x\times 10^{-12}m$. The value of $x$ is _______ (Nearest integer)
Given : Mass of electron $=9.1\times 10^{-31}kg$
Charge on an electron $=1.6\times 10^{-19}C$
Planck's constant $=6.63\times 10^{-34}JS$

Answer 1

Answer: 6

De-broglie-wave length of electron:
${\lambda }_e=\frac{h}{\sqrt{2m(KE)}}$
$\{\because e^{-}$is accelerated from rest $\Rightarrow KE=q\times V$
$\lambda =\frac{h}{\sqrt{2mqv}}$
$=\frac{6.63\times 10^{-34}}{\sqrt{2\times 1.6\times 10^{-19}\times 9.1\times 10^{-31}\times 40\times 10^3}}$
$=0.614\times 10^{-11}m$
$=6.14\times 10^{-12}m$
Nearest integer $=6$