Answer

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Hint: In this question use the concept that the total energy (T.E) is the sum of kinetic energy (K) and the potential energy (U), that is T.E = K + U. Then use the relation that kinetic energy (K) is the modulus of the total energy (T.E) and potential energy (U) is the double of the total energy (T.E) such that the total energy which is the sum of kinetic energy and potential energy is the same. This will help approaching the problem statement.

Given data:

Total energy of an electron in an excited state of hydrogen atom is -3.4eV

Now we have to find out the kinetic and potential energy of electrons in this state.

As we know that the total energy (T.E) is the sum of kinetic energy (K) and the potential energy (U)

Therefore, T.E = K + U

Now as we know that kinetic energy (K) is the modulus of the total energy (T.E).

$ \Rightarrow K = \left| {T.E} \right|$= |-3.4| = 3.4eV

As modulus of any negative quantity is positive.

And the potential energy (U) is the double of the total energy (T.E) so that the total energy which is the sum of kinetic energy and potential energy is the same.

$ \Rightarrow U = 2\left( {T.E} \right)$ = 2(-3.4) = -6.8eV

So the total energy, T.E = K + U = 3.4 + (-6.8) = -3.4eV which is the same as the above total energy.

So kinetic energy of electron in this state is, K = 3.4eV

And potential energy of electron in this state is, U = -6.8eV

So this is the required answer.

Hence option (B) is the correct answer.

Note – In quantum mechanics excited state can be considered as any quantum state that is at higher energy level from the ground. The ground state basically depicts the lowest energy levels that an atom can have.

__Complete step-by-step solution__-Given data:

Total energy of an electron in an excited state of hydrogen atom is -3.4eV

Now we have to find out the kinetic and potential energy of electrons in this state.

As we know that the total energy (T.E) is the sum of kinetic energy (K) and the potential energy (U)

Therefore, T.E = K + U

Now as we know that kinetic energy (K) is the modulus of the total energy (T.E).

$ \Rightarrow K = \left| {T.E} \right|$= |-3.4| = 3.4eV

As modulus of any negative quantity is positive.

And the potential energy (U) is the double of the total energy (T.E) so that the total energy which is the sum of kinetic energy and potential energy is the same.

$ \Rightarrow U = 2\left( {T.E} \right)$ = 2(-3.4) = -6.8eV

So the total energy, T.E = K + U = 3.4 + (-6.8) = -3.4eV which is the same as the above total energy.

So kinetic energy of electron in this state is, K = 3.4eV

And potential energy of electron in this state is, U = -6.8eV

So this is the required answer.

Hence option (B) is the correct answer.

Note – In quantum mechanics excited state can be considered as any quantum state that is at higher energy level from the ground. The ground state basically depicts the lowest energy levels that an atom can have.

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