The minimum excitation energy of an electron revolving in the first orbit of hydrogen is (in $eV$)

The ionization energy of hydrogen is $13.6 \text{ eV}$. The energy of the photon released when an electron jumps from the first excited state $(n=2)$ to the ground state of a hydrogen atom is (in $\text{ eV}$)

The energy of an electron in the first excited state of an $H$-atom is $-3.4 \ eV$. Its kinetic energy is ........ $eV$.

$A$ 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:

As the quantum number $n$ increases,the difference in energy between consecutive energy levels

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: