Kinetic energy of electron = E_k = 120 eV

Plancks constant = h = 6.6 x 10^-34 Js

Mass of an electron = 9.1 x 10^-31 kg

Charge on an electron = 1.6 x 10^-19 C

a) For an electron, the kinetic energy can be expressed as - 

(1/2) mv² = E_k

v² = 2eE_k/m

Therefore, v = √[(2 x 1.6 x 10^-19 x 120)/9.1 x 10^-31]

= 6.496 x 10⁶ m/s

The momentum of an accelerated electron = p = mv

So p = 9.1 x 10^-31 x 6.496 x 10⁶ 

= 5.91 x 10^-24 kg m/s

Thus, momentum of each electron is 5.91 x 10^-24 kg m/s.

b) Speed of electron = v = 6.496 x 10⁶ m/s.

c) De broglie wavelength of an electron with acceleratinf potential V is given by -

λ = h/p Angstorms

That is, 6.6 x 10^-34/5.91 x 10^-24 metres = 1.116 x 10^-10 metres

= 0.112 nm.

Therefore, the de broglie wavelength of each electron is 0.112 nm.