Ionization energy of hydrogen is `13.6 eV`. If `h = 6.6 xx 10^(-34) J-s`, the value of `R` will of the order of

Given that ionization energy of hydrogen atom is 13.6 eV, calculate the wavelength of first line of the Lyman series.

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

The ionization potential of hydrogen is -13.6 eV. The energy required to excite the electron from the first to the third orbit is approximately

The energy of a photon of wavelength 4000 Å is (use h=6.66xx10^(-34) J-s)

If ionization enthalpy for hydrogen atom is 13.6 eV, then ionization enthalpy for He^+ will be