Question

The first excited state of a particular atom in a gas is 6.1 eV above the ground state. A moving electron collides with one of these atoms, and excites the atom to its first excited state. Immediately after the collision the kinetic energy of the electron is 3.2 eV. What was the kinetic energy of the electron just before the collision

Answer 1

Answer:

The kinetic energy of the electron just before the collision is 9.3 eV.

Explanation:

We can find the kinetic energy of the electron before the collision can be found by energy conservation:

$E_{i} = E_{f}$

$K_{a_{i}} + K_{e_{i}} = K_{a_{f}} + K_{e_{f}}$   (1)

Where:

$K_{a_{i}}$: is the initial kinetic energy of the atom

$K_{a_{f}}$: is the final kinetic energy of the atom = 6.1 eV + $K_{a_{i}}$

$K_{e_{i}}$: is the initial kinetic energy of the electron =?

$K_{e_{f}}$: is the final kinetic energy of the electron = 3.2 eV

By solving equation (1) for $K_{e_{i}}$ we have:

$K_{a_{i}} + K_{e_{i}} = (6.1 eV + K_{a_{i}}) + 3.2 eV$

$K_{e_{i}} = 6.1 eV + 3.2 eV = 9.3 eV$

Therefore, the kinetic energy of the electron just before the collision is 9.3 eV.

I hope it helps you!