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Energy E of a hydrogen atom with princip...

Energy `E` of a hydrogen atom with principle quantum number `n` is given by `E = (-13.6)/(n^(2)) eV`. The energy of a photon ejected when the electron jumps from `n = 3` state to `n = 2` state of hydrogen is approximately

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Energy of a hydrogen atom with principal quantum number n is shown by E = (-13.6)/(n^(2)) eV . The energy of a photon ejected when the electron jumps from n = 3 state to n = 2 state of hydrogen is approximately.

Energy E of a hydrogen atom with principal quantum number n is given by E_n=-13.6/n^2 eV . The energy of photon ejected when the electron jumps from n=3 state to n=2 state of hydrogen is approximately

The ionisation energy of hydrogen is 13.6 eV . The energy of the photon released when an electron jumps from the first excited state (n=2) to the ground state of hydrogen atom is

The ionisation energy of hydrogen is 13.6 eV . The energy of the photon released when an electron jumps from the first excited state (n=2) to the ground state of hydrogen atom is

The ionisation energy of hydrogen is 13.6 eV. The energy of the photon released when an electron jumps from the first excited state (n=2) to the ground state of a hydrogen atom is