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

When ultraviolet rays of wavelength 620 Å is incident on a hydrogen atom in the ground state, its electron is emitted with a velocity of 1.5 xx10^(6) m *s^(-1) . What is the ionisation energy of hydrogen ? Given , h= 6.625xx10^(-34)J *s , mass of electron = 9.1 xx 10^(-31)kg, c=3xx 10^(8)m*s^(-1) and 1 eV=1.6xx 10^(-19)J .

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)