(a) what is the ionization energy of a hydrogen atom that is in the n = 6 excited state? (b) for a hydrogen atom, determine the ratio of the ionization energy for the n = 6 excited state to the ionization energy for the ground state.

(a) 0.3778 eV

(b) Ratio = 0.0278

Explanation:

The Bohr's formula for the calculation of the energy of the electron in nth orbit is:

(a) The energy of the electron in n= 6 excited state is:

Ionisation energy is the amount of this energy required to remove the electron. Thus, |E| = 0.3778 eV

(b) For first orbit energy is:

Ratio = 0.0278

this means that the temperature of both substances will be equal

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