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The ground state energy of a hydrogen atom is $ - 13.6eV$. What are the kinetic and potential energies of the electron in this state?

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Last updated date: 19th Apr 2024
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Answer
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Hint: From the given ground state energy, we can calculate the kinetic energy by using $KE = - E$, where KE is the kinetic energy and E is the ground state energy. And the potential energy will be given by using $PE = 2 \times KE$, where PE is the potential energy and KE is the kinetic energy.

Complete step-by-step answer:
Given ground state energy of hydrogen atom is E$ = - 13.67eV$
The ground state of a quantum- mechanical system is its lowest-energy state; the energy of the ground state is known as the zero-point energy of the system. An excited state is any state with energy greater than the ground state.
The kinetic energy of the electron is given as:
$KE = - E$,
Where, KE is the kinetic energy and E is the ground state energy.
Kinetic Energy: The Kinetic energy (KE) of an object is the energy that it possesses due to its motion. It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity. Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes
$\therefore KE = - \left( { - 13.67eV} \right)$
$\therefore KE = + 13.67eV$
The potential energy of the electron is given as:
$PE = - 2 \times KE$
Where PE is the potential energy and E is the ground state energy.
Potential Energy: Potential energy is the energy held by an object because of its position relative to other objects, stresses within itself, its electric charge, or other factors.
$PE = - 2 \times 13.67{\text{eV}}$
$\therefore PE = - 27.2eV$

Note: The ground state energy is the total energy. E= T+V where T and V are kinetic and potential energy of the system. Therefore adding the above values that is $13.67 + \left( { - 27.2} \right)$ gives the $ - 13.67eV$ which is the ground state energy (E).
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