91Ó°ÊÓ

The potential difference through the electron moves is, $â–³V=44V$.

The initial kinetic energy of the electron is, $K_i=0$.

When a charged particle moves in an electric field through a potential difference then the energy required to move the particle is described as the ‘electrical potential energy’.

The value of the electrical potential energy relies upon the particle’s charge and the potential difference between two locations.

The formula for the change in the electrical potential energy required to move the electron is given by,

$\begin{array}{l}â–³U_{eI}=-eâ–³V\\ â–³U_{eI}=-e\left(44V\right)\\ â–³U_{eI}=-44eV\end{array}$

Applying work-energy principle, the change in kinetic energy of the electron is given by,

$\begin{array}{l}â–³K_e=-â–³U_{eI}\\ â–³K_e=-\left(-44eV\right)\\ â–³K_e=44eV\end{array}$

The final kinetic energy of the electron is given by,

$$\begin{gathered} â–³K_e=K_f-K_i \\ 44eV=K_f-0 \\ {K}_f=44eV \end{gathered}$$

Hence, the final kinetic energy of the electron is 44 eV.

If the particle is a proton, then its charge value is positive. So, the value of the change in the electrical potential energy required to move the proton would also be positive.

Then, applying work-energy principle, the change in kinetic energy of the proton would be negative.

The direction of the electric field of a proton is directly opposite to the direction of the electric field of an electron. So, the proton moves in the opposite direction from the electron.

Hence, the correct options are (1) The kinetic energy of the proton would be negative, and (2) The proton would move in the opposite direction from the electron.