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The total energy of an electron in the first excited state of the hydrogen atom isabout –3.4 eV.What is the kinetic energy of the electron in this state?What is the potential energy of the electron in this state?

Question

The total energy of an electron in the first excited state of the hydrogen atom isabout –3.4 eV.What is the kinetic energy of the electron in this state?What is the potential energy of the electron in this state?

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Solution

The total energy of an electron in an atom is the sum of its kinetic and potential energy. In a hydrogen atom, the kinetic energy and potential energy are related by the equation:

PE = -2(KE)

This is because the potential energy is negative, due to the attractive force between the electron and the proton, and it is twice the magnitude of the kinetic energy.

Given that the total energy (TE) is the sum of the kinetic energy (KE) and the potential energy (PE), we can write:

TE = KE + PE

We know that the total energy is -3.4 eV. We can substitute PE in the above equation with -2(KE) to get:

-3.4 eV = KE - 2(KE)

Solving this equation gives:

KE = -3.4 eV / -3 = 1.13 eV

Substituting KE = 1.13 eV in the equation PE = -2(KE) gives:

PE = -2 * 1.13 eV = -2.26 eV

So, the kinetic energy of the electron in the first excited state of the hydrogen atom is about 1.13 eV, and the potential energy is about -2.26 eV.

This problem has been solved

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