Login/Sign Up
Home » Calculate the (a) momentum, and (b) the de Broglie wavelength of the electrons accelerated through a potential difference of .
Calculate the (a) momentum, and (b) the de Broglie wavelength of the electrons accelerated through a potential difference of 56 \mathrm{~V}.
CBSE | Class 12 | Dual Nature of Radiation and Matter | NCERT | Physics
Calculate the (a) momentum, and (b) the de Broglie wavelength of the electrons accelerated through a potential difference of 56 \mathrm{~V}.

Potential difference is given in the question as \mathbf{V}=\mathbf{5 6 V}

Planck’s constant, \mathbf{h}=6.6 \times 10^{-34} \mathrm{~J} \mathrm{~s}

Mass of an electron, \mathbf{m}=\mathbf{9 . 1 x} \mathbf{1 0}^{-31} \mathbf{K g}

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

(a) The kinetic energy of each electron is equal to the accelerating potential t equilibrium,so we can write the relation of velocity (v) of each electron as:

\frac{1}{2} m v^{2}=\mathrm{eV}

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=m v

=9.1 \times 10^{-31} \times 4.44 \times 10^{6}=4.04 \times 10^{-24} \mathrm{Kg} \mathrm{m} / \mathrm{s}

Therefore, the momentum of each electron is 4.04 \times 10^{-24} \mathrm{Kg} \mathrm{m} / \mathrm{s}.

(b) Through a potential \mathrm{V}, de Broglie wavelength of an electron accelerating is given by the relation:

\lambda=\frac{12.27}{\sqrt{V}} \mathrm{~A}^{\circ}

=\frac{12.27}{\sqrt{56}} \times 10^{-19} \mathrm{~m}=0.1639 \mathrm{~nm}

As a result, 0.1639 \mathrm{~nm} is the de Broglie wavelength of each electron.

i

More Material