The first excited state of a particular atom in a gas is 5.7 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 3.4 ev. what was the kinetic energy of the electron just before the collision?

Kinetic energy, E = 9.1 eV

Explanation:

It is given that,

At the first excited state of a particular atom is 5.7 eV above the ground state. The kinetic energy of the atom after the collision is 3.4 eV.

We need to find the kinetic energy of the electron just before the collision. The conservation of momentum will be followed here. The energy gets transferred from moving electrons to these atoms is 5.7.

Let E is the kinetic energy of the electron just before the collision. It is equal to :

E = 5.7 eV + 3.4 eV

E = 9.1 eV

So, the kinetic energy of the electron just before the collision is 9.1 eV. Hence, this is the required solution.

i just did an experiment like this in my science class. also the answer is yes, it will get separated because they are two opposites meaning if they are opposite they can not attract.

hope this .!

the ima (higher than 1m) will be greater than the ima with the height of 1m.