On a planet where the temperature is so high, the ground state of an electron in the hydrogen atom is n = 4. What is the ratio of ionization energy (IE) on this planet to that on earth?

The ratio of ionization energy (IE) on this planet to that on earth is 1:16.

The energy associated with the ground state of the hydrogen atom = - 13.6 ev/n^2

E =  - 13.6 ev/(1)^2 = -13.6 ev

Now, the ionization energy occurs at E0 hence;

0 - (-13.6 ev) = 13.6 ev

In that planet, the energy of the ground state = - 13.6 ev/(4)^2 = -0.85 ev

The ionization energy = 0 - ( -0.85 ev) = 0.85 ev

The ratio of the ionization energy is; 0.85 ev/ 13.6 ev = 1:16

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1:16

Explanation:

The ground state of an electron on the planet is n = 4 compared the ground state of an electron at n =1. For a hydrogen atom, the electron energy level is given as:

Hence the ratio of their ionization energies is 1:16