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

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

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

The wavelength of the first line of the Lyman series for hydrogen is $\lambda$. What is the wavelength of the second line of the Balmer series for a $He^+$ ion?

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

An ionic atom equivalent to hydrogen atom has wavelength equal to ${\rm{1/4}}$ of the wavelength of hydrogen lines. The ion will be

In a spectrum of hydrogen atom, the ratio of shortest wavelength in Lyman series to longest wavelength in Paschen series 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

The ratio of wavelengths of the last line of Balmer series and the last line of Lyman series is

The wavelength of Lyman series is

The ratio of wavelengths of the last line of Balmer series and the last line of Lyman series is