Question

Through what potential difference must a proton be accelerated to reach the speed it would have by falling $100m$in vacuum?

Short answer

The potential difference by which the proton is to be accelerated is$∆V=\mathbf{1}\mathbf{.}\mathbf{02}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathbf{}\mathit{v}\mathit{o}\mathit{l}\mathit{t}\mathit{s}$

Step-by-step solution

Given information and theory used 

Given : Proton falls : $100m$in vacuum

Theory used :

From the equation of conservation of energy, we have :$mgh=q∆V$

calculating the potential difference by which the proton is to be accelerated.

A proton will have an energy of $mgh$if it falls in vacuum from a height of $h$

Rewriting rule of conservation of energy we have :

$∆V=\frac{mgh}{q}$

In terms of numbers, we have

$∆V=\frac{1.67\times {10}^{-27}\times 9.81\times 100}{1.6\times {10}^{-19}}=\boxed{\mathbf{1}\mathbf{.}\mathbf{02}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}\mathbf{}\mathit{v}\mathit{o}\mathit{l}\mathit{t}\mathit{s}}$