Calculate the

**(a) ** momentum,
and

**(b) ** de Broglie
wavelength of the electrons accelerated through a potential
difference of 56 V.

Potential
difference, *V*
= 56 V

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)** At
equilibrium, the kinetic energy of each electron is equal to the
accelerating potential, i.e., we can write the relation for velocity
(*v*) of
each electron as:

The momentum of each accelerated electron is given as:

*p*
= *mv*

= 9.1 × 10^{−31}
× 4.44 × 10^{6}

= 4.04 × 10^{−24}
kg m s^{−1}

Therefore,
the momentum of each electron is 4.04 × 10^{−24}
kg m s^{−1}_{.}

**(b) **De Broglie
wavelength of an electron accelerating through a potential *V*,
is given by the relation:

_{
}

Therefore, the de Broglie wavelength of each electron is 0.1639 nm.

**
**