Excitation Energy Of Hydrogen

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A hydrogen atom initially in its ground state absorbs a photon and its electron jumps to the second excited state with principal quantum number three. How much energy must the absorbed photon carry for this transition?

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The energy a photon must supply equals the difference between the final and initial level energies, where each level is $E_n = -\dfrac{13.6}{n^2}$ eV. The ground state has $E_1 = -13.6$ eV and the second excited state has $E_3 = -\dfrac{13.6}{9} = -1.51$ eV. The required excitation energy is $\Delta E = E_3 - E_1 = -1.51 - (-13.6) = 12.09$ eV. The value 13.6 eV is the full ionisation energy needed to remove the electron entirely ($n = 1 	o \infty$), not to reach $n = 3$. The value 10.2 eV corresponds only to the $n = 1 	o 2$ transition, the first excited state. The value 1.51 eV is merely the magnitude of $E_3$ and is not the energy gap. Only photons whose energy matches an allowed gap can be absorbed; a photon carrying slightly more or less simply passes through the atom without exciting the electron, which is why atomic absorption spectra consist of sharp lines. This applies the NCERT energy-difference rule for photon absorption. A plausibility check confirms the answer lies just below the 13.6 eV ionisation energy, as expected for a transition that nearly, but not quite, frees the electron.

Excitation Energy Of Hydrogen

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