The energy required to ionise a hydrogen like ion in its ground state is 9 Rydbergs. What is the wavelength of the radiation emitted when the electron in this ion jumps from the second excited state to the ground state?

We have a hydrogen-like ion whose ionisation energy in the ground state is given as $$9$$ Rydbergs. For any hydrogen-like species the ionisation energy from the ground state is given by the formula

$$E_{\text{ion}} = Z^{2}R_{H},$$

where $$Z$$ is the atomic number (nuclear charge) and $$R_{H}$$ is the Rydberg constant expressed in energy units (1 Rydberg = 13.6 eV). Substituting the given value,

$$Z^{2}R_{H}=9\,R_{H}\;\;\Longrightarrow\;\;Z^{2}=9\;\;\Longrightarrow\;\;Z=3.$$

So the ion is $$\text{Li}^{2+}$$ (a 3-proton nucleus with one electron).

The energy of the $$n^{\text{th}}$$ Bohr orbit for any hydrogen-like ion is given by

$$E_{n}=-\dfrac{Z^{2}R_{H}}{n^{2}}.$$

The phrase “second excited state” means $$n=3$$ (ground state $$n=1$$, first excited $$n=2$$, second excited $$n=3$$). When the electron falls from this state to the ground state, the energy released is

$$\Delta E = E_{1}-E_{3}= \left(-\dfrac{Z^{2}R_{H}}{1^{2}}\right)-\left(-\dfrac{Z^{2}R_{H}}{3^{2}}\right)=Z^{2}R_{H}\left(1-\dfrac{1}{9}\right)=Z^{2}R_{H}\left(\dfrac{8}{9}\right).$$

With $$Z^{2}=9$$ we get

$$\Delta E = 9R_{H}\left(\dfrac{8}{9}\right)=8R_{H}.$$

Because $$1\,\text{Rydberg}=13.6\,\text{eV},$$ the photon energy in electron-volts is

$$\Delta E = 8 \times 13.6\,\text{eV}=108.8\,\text{eV}.$$

To convert this energy into wavelength we use the relation

$$E=\dfrac{hc}{\lambda}\quad\Longrightarrow\quad \lambda=\dfrac{hc}{E}.$$

In convenient units, $$hc=1240\,\text{eV nm}.$$ Substituting the numerical values,

$$\lambda=\dfrac{1240\,\text{eV nm}}{108.8\,\text{eV}}\approx11.4\,\text{nm}.$$

Hence, the correct answer is Option B.