What is the: (a) Momentum, (b) Speed of an electron with a kinetic energy of $120 \mathrm{eV} .$

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What is the: (a) Momentum, (b) Speed of an electron with a kinetic energy of $120 \mathrm{eV} .$

Kinetic energy of the electron is given as $\mathbf{E}_{\mathrm{K}}=\mathbf{1 2 0} \mathrm{eV}$

Planck’s constant, $\mathbf{h}=\mathbf{6 . 6} \times \mathbf{1 0}^{-34} \mathrm{~J} \mathrm{~s}$

Mass of an electron, $\mathbf{m}=\mathbf{9 . 1} \times \mathbf{1 0}^{-31} \mathrm{Kg}$

Charge on an electron, $\mathbf{e}=\mathbf{1 . 6} \times \mathbf{1 0}^{-19} \mathbf{C}$

(a) For an electron, the relation for kinetic energy can be wrtitten as,

$\mathrm{E}_{\mathrm{k}}=\frac{1}{2} m v^{2}$

Where, $v=$ speed of the electron

Therefore, $v^{2}=\sqrt{\frac{2 e E_{k}}{m}}$

$=\sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 120}{9.1 \times 10^{-31}}}$

$=\sqrt{42.198 \times 10^{12}}$

$=6.496 \times 10^{6} \mathrm{~m} / \mathrm{s}$

Momentum of the electron is,
$p=m v=9.1 \times 10^{-31} \times 6.496 \times 10^{6}$

$=5.91 \times 10^{-24} \mathrm{~kg} \mathrm{~m} / \mathrm{s}$

Therefore, $5.91 \times 10^{-24} \mathrm{Kg} \mathrm{m} / \mathrm{s}$ is the momentum of the electron.

(b) speed of the electron, $v=6.496 \times 10^{6} \mathrm{~m} / \mathrm{s}$

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