Question

What is the speed of a proton that has been accelerated from rest through a potential difference of $-1000\mathrm{V}$ ?

Short answer

The proton speed is$4.38\times {10}^5\mathrm{m}/\mathrm{s}$.

Step-by-step solution

Given information

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

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

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

Formulae used

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.

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

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

$W=qV$

Find the speed of proton

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,

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

Creating an equation for electron velocity v

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

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

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

$\begin{array}{rcl}v& =& \sqrt{\frac{2\left(1.6\times {10}^{-19}\mathrm{C}\right)(1000\mathrm{V})}{\left(1.67\times {10}^{-27}\mathrm{kg}\right)}}\\ & =& 4.38\times {10}^5{\mathrm{ms}}^{-1}\end{array}$

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