The total energy of an electron in the first excited state of the hydrogen atom is about -3.4 eV. what is the kinetic energy of the electron in this state ?
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the ground state of hydrogen atom is at energy -13.6 eV.
naturally this energy value is total energy.
E = T + V where T and V ar kinetic and potential energy of the system.
This energy is the eigen value of the Hamiltonian operator in ground state.
Therefore T = E -V ; as V = -2.T
will be E = -13.6 + 2.T therefore T = +(13.6 eV)
therefore the K.E. = 13.6 eV and potential energy will be - -27.2 eV so that the ground state is at -13.6 eV.
Hope it helps you
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naturally this energy value is total energy.
E = T + V where T and V ar kinetic and potential energy of the system.
This energy is the eigen value of the Hamiltonian operator in ground state.
Therefore T = E -V ; as V = -2.T
will be E = -13.6 + 2.T therefore T = +(13.6 eV)
therefore the K.E. = 13.6 eV and potential energy will be - -27.2 eV so that the ground state is at -13.6 eV.
Hope it helps you
Please make me as brainliest
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