Electric Potential — Question 2

Question 2

An ion gains several electronvolts per volt of potential drop

Use $e=1.602176634\times10^{-19}$ C, exact. An ion of charge $+3e$ starts from rest and crosses a potential drop of 120 V. Only the electric force does work.

1. Find its final kinetic energy in electronvolts and joules.

2. Find the potential drop required for the same ion to gain 1.00 keV, and distinguish voltage from energy.

Question 2 - Solution

An ion gains several electronvolts per volt of potential drop

(a) Method and calculation

$$K_f=-q\Delta V=-(3e)(-120\,\mathrm V)=360\,\mathrm{eV}.$$$$K_f=360(1.602176634\times10^{-19})=5.76784\times10^{-17}\,\mathrm J.$$

(b) Interpretation and further calculation

$$|\Delta V|=\frac{1000\,\mathrm{eV}}{3e}=333.333\,\mathrm V.$$$$1\,\mathrm{eV}=e(1\,\mathrm V),\qquad 1\,\mathrm V=1\,\mathrm{J/C}.$$

Physical check

The numerical energy gain in eV equals the voltage drop multiplied by the positive ion charge number.