Question

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

Answer 1

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.