Question:
Calculate the (a) momentum, and (b) de Broglie wavelength of the electrons accelerated through a potential difference of 56 V.
Answer:
Potential difference V = 56 V
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) At equilibrium condition, the kinetic energy of an electron = its acclerating potential. Therefore, its velocity can be expressed as-
(1/2) mv2 = eV
v2 = 2eV/m
Therefore, v = √[(2 x 1.6 x 10-19 x 56)/9.1 x 10-31]
= 4.44 x 106 m/s
The momentum of an accelerated electron = p = mv
So p = 9.1 x 10-31 x 4.44 x 106
= 4.04 x 10-24 kg m/s
Thus, momentum of each electron is 4.04 x 10-24 kg m/s.
b) De broglie wavelength of an electron with acceleratinf potential V is given by -
λ = 12.27/√V Angstorms
That is, (12.27/√56) x 10-10 metres
= 0.1639 nm.
Therefore, the de broglie wavelength of each electron is 0.1639 nm.
Not what you are looking for? Go ahead and submit the question, we will get back to you.
