Considering that a quintuply ionized carbon ion, C^{5+}, behaves like a hydrogen atom, calculate
(a) the radius r_n and energy E_n for a given state n and compare them with the corresponding expressions for hydrogen,
(b) the ionization energy of C^{5+} when it is in its first excited state and compare it with the corresponding value for hydrogen, and
(c) the wavelength corresponding to the transition from state n = 3 to state n = 1; compare it with the corresponding value for hydrogen.
(a) The C^{5+} ion is generated by removing five electrons from the carbon atom. To find the expressions for r_{n_C} and E_{n_C} for the C^{5+} ion (which has 6 protons), we need simply to insert Z = 6 into (1.76) r_n=\left(1+\frac{m_e}{M} \right)\frac{a_0}{Z}n^2, E_n=-\frac{Z^2}{1+m_e/M}\frac{R}{n^2}, :
r_{n_C}=\frac{a_0}{6}n^2, E_{n_C}=-\frac{36R}{n^2}, (1.193)
where we have dropped the term m_e/M, since it is too small compared to one. Clearly, these expressions are related to their hydrogen counterparts by
r_{n_C}=\frac{a_0}{6}n^2=\frac{r_{n_H}}{6} , E_{n_C}=-\frac{36R}{n^2}=36E_{n_H}. (1.194)
(b) The ionization energy is the one needed to remove the only remaining electron of the C^{5+} ion. When the C^{5+} ion is in its first excited state, the ionization energy is
E_{2_C}=-\frac{36R}{4}=-9\times 13.6 eV=-122.4 eV, (1.195)
which is equal to 36 times the energy needed to ionize the hydrogen atom in its first excited state:
E_{2_H}=-3.4 eV (note that we have taken n = 2 to correspond to the first excited state; as a result, the cases n = 1 and n = 3 will correspond to the ground and second excited states, respectively).
(c) The wavelength corresponding to the transition from state n = 3 to state n = 1 can be inferred from the relation hc/\lambda =E_{3_C}-E_{1_C} which, when combined with E_{1_C}=-489.6 eV and E_{3_C}=-54.4 eV, leads to
\lambda =\frac{hc}{E_{3_C}-E_{1_C}}=\frac{2\pi \hbar c}{E_{3_C}-E_{1_C}} =\frac{2\pi 197.33\times 10^{-9} eV m}{-54.4 eV+489.6 eV}=2.85 nm. (1.196)