Energy of an electron (in $eV$) in the $2^{nd}$ orbit of a $He^{+}$ ion is (in $.6$)?

If $\lambda_1$ and $\lambda_2$ are the wavelengths of the photons emitted when electrons in the $n^{\text{th}}$ orbit of a hydrogen atom fall to the first excited state and ground state respectively,then the value of $n$ is:

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 $E$?

In a hydrogen-like atom,when an electron jumps from the $M$-shell to the $L$-shell,the wavelength of the emitted radiation is $\lambda$. If an electron jumps from the $N$-shell to the $L$-shell,the wavelength of the emitted radiation will be:

The ionisation potential of hydrogen atom is $13.6 \ V$. The energy required to remove an electron from the second orbit of hydrogen is......$eV$.

$A$ 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: