Question

The first excited state of a mercury atom is 4.9eV above the ground state. A moving electron collides with a mercury atom and excites the mercury atom to its first excited state. Immediately after the collision the kinetic energy of the electron is 0.3eV. What was the kinetic energy of the electron just before the collision?

Short answer

The kinetic energy of the electron just before the collision was 5.2 eV.

Step-by-step solution

Definition of Law of conservation of energy

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 $e_f=0.3eV$.

Application of energy conservation equation

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$5.2eV$ .