Final answer:
The wavelength of the radiation emitted when an electron in a hydrogen-like ion jumps from the second excited state to the ground state can be calculated using the Rydberg formula for hydrogen-like ions, considering the energy difference between the two states. The atomic number of the ion is needed for the exact calculation.
Step-by-step explanation:
The question is asking to calculate the wavelength of the radiation emitted when the electron in a hydrogen-like ion jumps from the second excited state to the ground state given that the energy needed to ionize the ion in its ground state is 9 Rydbergs. Using the Rydberg formula for hydrogen-like ions and the given ionization energy, the wavelength can be calculated using the energy difference between the initial (n=3 for second excited state) and final state (n=1 for ground state).
The energy levels of hydrogen-like ions are given by En = -Z2R∞/n2, where Z is the atomic number, R∞ is the Rydberg constant for hydrogen (13.6 eV), and n is the principal quantum number.
The energy difference ΔE between the ground state and the second excited state (ΔE = E1 - E3), multiplied by the Planck's constant (h) and the speed of light (c), gives the photon energy. The wavelength (λ) is then found by λ = hc/ΔE. Be aware that the specific atomic number (Z) of the ion is needed to calculate the actual wavelengths, but the concept remains the same.