Calculate the

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, $h=6.6 \times 10^{-34} \mathrm{Js}$

Mass of an electron, $m=9.1 \times 10^{-31} \mathrm{~kg}$

Charge on an electron, $e=1.6 \times 10^{-19} \mathrm{C}$

(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:

$\frac{1}{2} m v^{2}=e V$

$v^{2}=\frac{2 e V}{m}$

$\therefore v=\sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 56}{9.1 \times 10^{-31}}}$

$=\sqrt{19.69 \times 10^{12}}=4.44 \times 10^{6} \mathrm{~m} / \mathrm{s}$

The momentum of each accelerated electron is given as:

p = mv

$=9.1 \times 10^{-31} \times 4.44 \times 10^{6}$

$=4.04 \times 10^{-24} \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}$

Therefore, the momentum of each electron is $4.04 \times 10^{-24} \mathrm{~kg} \mathrm{~m} \mathrm{~s}^{-1}$

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

$\lambda=\frac{12.27}{\sqrt{V}} \AA$

$=\frac{12.27}{\sqrt{56}} \times 10^{-10} \mathrm{~m}$

$=0.1639 \mathrm{~nm}$

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


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