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?

Question

Juan Rose

Physics

11 April, 10:11

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?

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Answers (1)

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Lillie Peterson · Answer 1

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.