If the ionisation energy for the hydrogen atom is $13.6 \ eV$,then the energy required to excite it from the ground state to the next higher state is nearly: (in $eV$)

$A$ hydrogen atom falls from $n^{\text{th}}$ higher energy orbit to the first energy orbit $(n=1)$. The energy released is equal to $12.75 \text{ eV}$. The $n^{\text{th}}$ orbit is:

The energy levels of a certain atom are $A, B,$ and $C$ in increasing order of energy,i.e.,$E_A < E_B < E_C$. If $\lambda_1, \lambda_2,$ and $\lambda_3$ are the wavelengths of radiation corresponding to transitions from $C$ to $B$,$B$ to $A$,and $C$ to $A$ respectively,which of the following relations is correct?

The energy levels of a hydrogen atom are shown below. The transition corresponding to the emission of the shortest wavelength is

The ground state energy of a hydrogen atom is $-13.6 \ eV$. The potential energy of the electron in the first excited state of hydrogen is (in $eV$)

The energy of a hydrogen atom in the $n^{th}$ orbit is $E_n$. What will be the energy in the $n^{th}$ orbit of a singly ionized helium atom?