Question

What potential difference is needed to accelerate a he+ ion (charge +e, mass 4u) from rest to a speed of 1.0×106 m/s ? express your answer using two significant figures?

Answer 1

For the law of conservation of energy, the loss in potential energy of the He+ ion should be equal to the gain in kinetic energy:
$-\Delta U=\Delta K$
which can be rewritten as
$-q \Delta V = \frac{1}{2}mv^2$
where
$q=+e = 1.6 \cdot 10^{-19}C$ is the charge of the ion,
$m=4u=4\cdot 1.67 \cdot 10^{-27}kg$ is the mass of the ion,
$v=1.0 \cdot 10^6 m/s$ is the speed of the ion.
By using these values, we find the potential difference needed:
$\Delta V = \frac{1}{2} \frac{mv^2}{-q}= \frac{1}{2} \frac{(4\cdot 1.67 \cdot 10^{-27}kg)(1.0 \cdot 10^6 m/s)^2}{-1.6 \cdot 10^{-19}C}= -20875 V=-21 kV$
and the negative sign means the final point is at lower voltage than the initial point, and this is correct, because the ion has positive charge and a positive charge travels naturally from higher voltages to lower voltages.

Answer 2

The potential difference required to accelerate the helium ion to a speed of   $1.0 \times {10^6}\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}$ is $\boxed{21\,{\text{kV}}}$ or $\boxed{20875\,{\text{V}}}$.

Further Explanation:

The potential difference, through which the helium ion passes, provides the potential energy and this potential energy provided by the potential difference to the Helium ion leads to the increase in the kinetic energy of the ion.

The charge on the $H{e^ + }$ ion is equal to the charge of one electron and the mass of the helium atom is equal to the mass of $4\,{\text{amu}}$.

The expression for the conservation of the energy of the Helium ion in the potential difference is:

$\Delta PE = \Delta KE$  

Substitute $q\Delta V$ for $\Delta PE$ and $\dfrac{1}{2}m{v^2}$ for $\Delta KE$ in above expression.

$\begin{aligned}q\Delta V &= \frac{1}{2}m{v^2} \hfill \\\Delta V &= \frac{{m{v^2}}}{{2q}} \hfill \\\end{aligned}$  

Here, $\Delta V$ is the potential difference, $m$ is the mass of the ion, $q$ is the charge over the ion.

Substitute $1.6 \times {10^{ - 19}}\,{\text{C}}$ for $q$, $\left( {4 \times 1.67 \times {{10}^{ - 27}}\,{\text{kg}}} \right)$ for $m$ and $1.0 \times {10^6}\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}$ for $v$ in above expression.

$\begin{aligned}\Delta V &= \frac{{\left( {4 \times 1.67 \times {{10}^{ - 27}}\,{\text{kg}}} \right){{\left( {1.0 \times {{10}^6}\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}} \right)}^2}}}{{2\left( {1.6 \times {{10}^{ - 19}}\,{\text{C}}} \right)}} \\&= \frac{{6.68 \times {{10}^{ - 15}}}}{{3.2 \times {{10}^{ - 19}}}}\,{\text{V}} \\&= 2{\text{0875}}\,{\text{V}} \\&\approx {\text{21}}\,{\text{kV}}\\\end{aligned}$  

Thus, the potential difference required to accelerate the helium ion to a speed of $1.0 \times {10^6}\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}$ is $\boxed{21\,{\text{kV}}}$ or $\boxed{20875\,{\text{V}}}$.

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Answer Details:

Grade: College

Chapter: Electrostatics

Subject: Physics

Keywords:  Potential energy, potential difference, accelerate, kinetic energy, speed, helium ion, charge, significant figures, rest to a speed.