91Ó°ÊÓ

The law of conservation of energy states that the total energy of an isolated system remains constant.

Assume the mercury atom and colliding electron to be an isolated system.

Then according to the law of conservation of energy, the total energy of the mercury atom (m) and colliding electron (e) system is conserved before and after the collision.

The initial energy of the mercury atom is zero and let the initial kinetic energy of the electron be$e_i$ .

The final energy of the mercury atom in its first excited state is$m_f=4.9eV$ and the final kinetic energy of the electron is $e_f=0.3eV$.

Apply the law of conservation of energy on the mercury atom and electron system.

$$\begin{gathered} m_i+e_i=m_f+e_f \\ 0+e_i=4.9eV+0.3eV \\ {e}_i=5.2eV \end{gathered}$$

Thus, the initial kinetic energy of the electron just before the collision was$5.2eV$ .