$A$ hydrogen atom makes a transition from $n = 2$ to $n = 1$ and emits a photon. This photon strikes a doubly ionized lithium atom $(Z = 3)$ in an excited state and completely removes the orbiting electron. The least quantum number for the excited state of the ion for this process is:

The minimum energy required to excite a hydrogen atom from its ground state is.....$ eV$.

When an electron in a hydrogen atom jumps from the third orbit to the second orbit,it emits a photon of wavelength $\lambda$. When it jumps from the fourth orbit to the third orbit,the wavelength emitted by the photon will be

The energy levels of an atom are shown below. When an electron makes a transition from an energy level of $2E$ to an energy level of $E$,the wavelength of the emitted electromagnetic radiation is $\lambda$. What is the wavelength of the emitted photon when the electron makes a transition from an energy level of $4E/3$ to an energy level of $E$?

Due to transitions among its first three energy levels,a hydrogenic atom emits radiation at three discrete wavelengths $\lambda_1, \lambda_2$ and $\lambda_3$ (where $\lambda_1 < \lambda_2 < \lambda_3$). Then,

The energy required to remove an electron from the first excited state of $Li^{++}$ is ......... $eV$.