What is the

**(a) ** momentum,

**(b) ** speed, and

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

Kinetic
energy of the electron, *E*_{k}
= 120 eV

Planck’s
constant, *h*
= 6.6 × 10^{−34}
Js

Mass
of an electron, *m*
= 9.1 × 10^{−31}
kg

Charge
on an electron, *e*
= 1.6 × 10^{−19}
C

**(a)** For
the electron, we can write the relation for kinetic energy as:

Where,

*v*
= Speed of the electron

Momentum
of the electron, *p*
= *mv*

= 9.1
× 10^{−31}
× 6.496 × 10^{6}

= 5.91
× 10^{−24}
kg m s^{−1}

Therefore,
the momentum of the electron is 5.91 × 10^{−24}
kg m s^{−1}_{.}

**(b)** Speed
of the electron, *v*
= 6.496 × 10^{6}
m/s

**(c)** De
Broglie wavelength of an electron having a momentum *p*,
is given as:

Therefore, the de Broglie wavelength of the electron is 0.112 nm.

**
**