Work required to pull an electron from helium atom is 40 eV. Then the work required to pull both the electrons from the helium atom is

If wavelength corresponding to 2nd line of Lyman series is $\lambda$, then the wavelength corresponding to last line of Balmer series will be

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

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

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

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 wavelength of the energy emitted when electron comes from fourth orbit to second orbit in hydrogen is $20.397\,\,cm$. The wavelength of energy for the same transition in $H{e^ + }$ is

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

The wavelength of radiation emitted is ${\lambda _0}$ when an electron jumps from the third to second orbit of hydrogen atom. For the electron jump from fourth to the second orbit of the hydrogen atom, the wavelength of radiation emitted will be

If ${\lambda _{\max }}$ is $6563\,\,{A^0}$, then wavelength of second line for Balmer series will be

The ratio of longest wavelength and the shortest wavelength observed in the five spectral series of emission spectrum of hydrogen is