Question

A completely ionized beryllium atom (net charge = +4e) is accelerated through a potential difference of 6.0 v. what is the increase in kinetic energy of the atom?

Answer 1

For the principle of conservation of energy, the increase in kinetic energy of the atom is equal to the energy lost when moving through the potential difference:
$\Delta K=\Delta U=q \Delta V$
where q is the charge of the Beryllium atom:
$q=4 e=4 \cdot (1.6 \cdot 10^{-19}C)=6.4 \cdot 10^{-19}C$
and $\Delta V=6.0 V$. Therefore, the increase in kinetic energy is
$\Delta K = (6.4 \cdot 10^{-19}C)(6.0 V)=3.8\cdot 10^{-18} J$