The first excited state of a particular atom in a gas is 5.2 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 2.2 eV. What was the kinetic energy of the electron just before the collision?

The initial kinetic energy of the electron is 7.4 eV.

The ground state is the lowest energy state of the atom. Other higher energy  states are called excited states. An atom can receive energy and move from ground state to excited state.

In this case; Kinetic energy lost by the electron = energy used to excite the atom = 5.2 eV

Kinetic energy of the electrons after collision = 2.2 eV

Hence, kinetic energy of the electron before collision =  5.2 eV + 2.2 eV = 7.4 eV

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7.4eV

Explanation:

We are given tha

Energy of  a particular atom in a gas in the first excited state=5.2 eV

Kinetic energy of the electron after collision=K.E=2.2 eV

We have to find the kinetic energy of the electron just before the collision.

We know that energy transferred from one particle to other atom.

By conservation law of energy

Kinetic energy of electron just before the collision=K.E of electron after collision+ energy of atom in a gas in the first excited state

Kinetic energy of electron just before the collision=2.2e V+5.2 eV=7.4eV

Hence, the kinetic energy of electron just before the collision=7.4 eV

Kalium Argon dating

Kalium or Potassium radioactive atom that commonly found in stone or artifact. Since Kalium is radioactive, it will be reduced and converted into another atom(in this case argon) after a set of time. You can determine the time elapsed by comparing the Kalium with argon mass.