Answer:
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:
[tex]E(n)=\frac{13.6}{n^2}\\ \\Where\ n\ is\ the\ ionization \ energy[/tex]
[tex]E(4)=\frac{13.6}{4^2}=\frac{13.6}{16}\\ \\E(1)=\frac{13.6}{1^2}=\frac{13.6}{1}\\ \\The\ ratio\ of\ their\ ionization\ energy=\frac{E(4)}{E(1)} =\frac{\frac{13.6}{16} }{\frac{13.6}{1} } =\frac{1}{16}=1:16[/tex]
Hence the ratio of their ionization energies is 1:16
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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