91Ó°ÊÓ

Potential difference, $V=-1000\mathrm{V}$

Mass of proton,$m_p=1.67×{10}^{-27}\mathrm{kg}$

Charge of proton, $q=1.6×{10}^{-19}\mathrm{C}$

Starting at rest, a proton of mass m acquires a charge q and a velocity v by applying an electric potential V. The relationship gives the change in kinetic energy of a proton.

$\begin{array}{rcl}\mathrm{Δ}K& =& \frac{1}{2}m{v}^2-0\\ & =& \frac{1}{2}m{v}^2\end{array}$

The work done by an electron's applied electric potential is given by

$W=qV$

The change in kinetic energy of an electron is quantitatively equivalent to the work done by the electric potential on an electron, according to the work energy theorem. Mathematically,

$\begin{array}{rcl}W& =& \mathrm{Δ}KqV\\ & =& \frac{1}{2}m{v}^2\end{array}$

Creating an equation for electron velocity v

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

Where, m is mass of the proton $\left(1.67×{10}^{-27}\mathrm{kg}\right)$and q is the charge on the proton $\left(1.6×{10}^{-19}\mathrm{C}\right)$.

Insert $1000\mathrm{V}$for the accelerating potential is $V,1.67×{10}^{-27}\mathrm{kg}$for role="math" localid="1649411463161" $s=1.6×{10}^{-19}\mathrm{C}$for q in the above equation

Thus, the proton speed is $4.38×{10}^5{\mathrm{ms}}^{-1}$