The energy required to remove the electron from a singly ionized Helium atom is 2.2 times the energy required to remove an electron from Helium atom. The total energy required to ionize the Helium atom completely is:

Excitation energy of a hydrogen like ion in its first excitation state is $40.8\,\,eV$. Energy needed to remove the electron from the ion in ground state is

The energy of electron in the nth orbit of hydrogen atom is expressed as${E_n} = \frac{{ - 13.6}}{{{n^2}}}eV$. The shortest and longest wavelength of Lyman series will be

In hydrogen atom, if the difference in the energy of the electron in $n = 2$ and $n = 3$ orbits is $E$, the ionization energy of hydrogen atom is

In terms of Rydberg’s constant $R$, the wave number of the first Balmer line is

An electron jumps from the ${4^{{\rm{th}}}}$ orbit to the ${2^{{\rm{nd}}}}$ orbit of hydrogen atom. Given the Rydberg’s constant $R = {10^5}\,\,c{m^{ - 1}}$. The frequency in $Hz$ of the emitted radiation will be

The wavelength of the first line of Balmer series is $6563\,\,{A^0}$. The Rydberg constant for hydrogen is about

The shortest wavelength in the Lyman series of hydrogen spectrum is 912\mathop {\rm{A}}\limits^ \circ   corresponding to a photon energy of $13.6\,\,eV$. The shortest wavelength in the Balmer series is about

The energy of a hydrogen atom in its ground state is $- 13.6\,\,eV$. The energy of the level corresponding to the quantum number $n = 2$ (first excited state) in the hydrogen atom is

In a hydrogen atom, which of the following electronic transitions would involve the maximum energy change

Taking Rydberg’s constant ${R_H} = 1.097\,\, imes \,\,{10^7}\,\,m$, first and second wavelength of Balmer series in hydrogen spectrum is