Answer
See explanation.
Work Step by Step
For large values of n, the inner electrons almost completely shield the nucleus. The effective atomic number is 1, and the ionization energy is $\frac{13.6eV}{n^2}$.
b. The ionization energy is $\frac{13.6eV}{300^2}=1.51\times10^{-4}eV$.
In the Bohr model, $r_n=n^2a$. Calculate the orbital radius.
$$r_{300}=300^2(0.529\times10^{-10}m)=4.76\times10^{-6}m$$
c. Repeat the calculation. The ionization energy is $\frac{13.6eV}{600^2}=3.78\times10^{-5}eV$.
Calculate the orbital radius.
$$r_{600}=600^2(0.529\times10^{-10}m)=1.90\times10^{-5}m$$
