Question

What is the

(a) momentum,

(b)speed, and

(c) de broglie wavelength of an electron with kinetic energy of 120 eV.

Answer 1

(a)

$E=120eV$

As, $m{v}^2/2=E;m$ is mass of electron.

So, $v=6.496\times {10}^{6}m/s$

momentum of electron $p=mv=5.91\times {10}^{-24}kgm/s$

(b)Forspeed, $m{v}^2/2=E\Rightarrow v=6.496\times {10}^{6}m/s$

(c)DeBroglie wavelength of an electron having momentum $p=\sqrt{2mE};E$is the Kinetic energy, is given by, $\lambda =h/p$.

Inthis case, $\lambda =0.112nm$.