The ionization energy of the hydrogen atom is $13.6 \ eV$. The potential energy of the electron in $n=2$ state of the hydrogen atom is

The diagram shows the energy levels for an electron in a certain atom. Which transition shown represents the emission of a photon with the most energy?

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,

As per the given figure,$A$,$B$,and $C$ are the first,second,and third excited energy levels of a hydrogen atom,respectively. If the ratio of the two wavelengths (i.e.,$\frac{\lambda_1}{\lambda_2}$) is $\frac{7}{4n}$,then the value of $n$ will be

With the increase in principal quantum number,the energy difference between two successive energy levels

If the electron in a hydrogen atom moves from the ground state orbit to the $5^{\text{th}}$ orbit,then the potential energy of the electron