Given a classical model of the tritium atom with a nucleus of charge $+1$ and a single electron in a circular orbit of radius $r_{0}$, suddenly the nucleus emits a negatron and changes to charge $+2$.
(The emitted negatron escapes rapidly and we can forget about it.) The electron in orbit suddenly has a new situation.

Find the ratio of the electron's energy after to before the emission of the negatron (taking the zero of energy, as usual, to be for zero kinetic energy at infinite distance).
As the negatron leaves the system rapidly, we can assume that its leaving has no effect on the position and kinetic energy of the orbiting electron.

From the force relation for the electron,

$$
\frac{m v_{0}^{2}}{r_{0}}=\frac{e^{2}}{4 \pi \varepsilon_{0} r_{0}^{2}},
$$

we find its kinetic energy

$$
\frac{m v_{0}^{2}}{2}=\frac{e^{2}}{8 \pi \varepsilon_{0} r_{0}}
$$

and its total mechanical energy

$$
E_{1}=\frac{m v_{0}^{2}}{2}-\frac{e^{2}}{4 \pi \varepsilon_{0} r_{0}}=\frac{-e^{2}}{8 \pi \varepsilon_{0} r_{0}}
$$

before the emission of the negatron. After the emission the kinetic energy of the electron is still $\frac{e^{2}}{8 \pi \varepsilon_{0} r_{0}}$, while its potential energy suddenly changes to

$$
\frac{-2 e^{2}}{4 \pi \varepsilon_{0} r_{0}}=-\frac{-e^{2}}{2 \pi \varepsilon_{0} r_{0}} .
$$

Thus after the emission the total mechanical energy of the orbiting electron is

$$
E_{2}=\frac{m v_{0}^{2}}{2}-\frac{2 e^{2}}{4 \pi \varepsilon_{0} r_{0}}=\frac{-3 e^{2}}{8 \pi \varepsilon_{0} r_{0}}
$$

giving

$$
\frac{E_{2}}{E_{1}}=3 .
$$

In other words, the total energy of the orbiting electron after the emission is three times as large as that before the emission.