What is the speed of an electron that has been accelerated from rest through a potential difference of 1050?

Solutions

Expert Solution

Applying the law of conservation of energy,

KE of electron = Potential energy

$$ \frac{1}{2} m\left(v^{2}-u^{2}\right)=\mathrm{qV} $$

Where $m=$ mass of elcetron, $u=$ initial speed $=0 m / s, v=$ final speed,

$\mathrm{q}=$ charge on electron, $\mathrm{V}=$ Potential diference $=1050$

$\mathrm{v}=\sqrt{\frac{2 \mathrm{qV}}{m}}$

Plugging the values,

$\mathrm{v}=\sqrt{\frac{2\left(1.6 \times 10^{-19}\right)(1050)}{9.1 \times 10^{-31}}}=\sqrt{369 \times 10^{12}}=19.21 \times 10^{6} \mathrm{~m} / \mathrm{s}$

Velocity of electron $=19.21 \times 10^{6} \mathrm{~m} / \mathrm{s}$ approx.

Dr. OWL answered 3 years ago

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