Question

What potential difference is needed to accelerate a $\mathrm{He}+$ion (charge $+e,\mathrm{mass}4u$) from rest to a speed of$2.0\times {10}^{6}m/s$?

Short answer

$∆V=\mathbf{8}\mathbf{.}\mathbf{35}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathbf{}\mathit{V}$

Step-by-step solution

Given information and theory used  

Given :

Element : $\mathrm{He}+$ion (charge $+e,\mathrm{mass}4u$)

To accelerate to a speed of : $2.0\times {10}^{6}m/s$

Theory used :

From the equation of conservation of energy, we have :

$q∆V=\frac{m}{2}({v_f}^2-{v_i}^2)$

Calculating potential difference needed  

Because all of the energy from the electric field is changed to kinetic energy, we get :

$q∆V=\frac{m{v_f}^2}{2}$

Taking into account the fact that there is no beginning velocity, we will have that. We may now write the solution as

$∆V=\boxed{\frac{m{v_f}^2}{2q}}$.

The potential difference can be calculated as $∆V=\frac{m{v_2}^2}{2q}$because the initial velocity is zero.

We get $∆V$by substituting the supplied numerical values :

$$\begin{gathered} ∆V=\frac{4\times 1.67\times {10}^{-27}{(2\times {10}^6)}^2}{2\times 1.6\times {10}^{-19}} \\ ∆V=\boxed{\mathbf{8}\mathbf{.}\mathbf{35}\mathbf{\times }{\mathbf{10}}^{\mathbf{4}}\mathbf{}\mathit{V}} \end{gathered}$$