An excited He^(+) ion emits two photons in succession, with wavelength 108.5 nm and 30.4 nm, in making a transition to ground state. The quantum number n, corresponding to its initial excited state is (for photon of wavelength lamda , energy E=(1240eV)/(lamda("in nm"))
A hydrogen atom ia in excited state of principal quantum number n . It emits a photon of wavelength lambda when it returnesto the ground state. The value of n is
Find the quantum number n corresponding to nth excited state of He^(++) ion if on transition to the ground state the ion emits two photons in succession with wavelength 108.5 nm and 30.4 nm. The ionization energy of the hydrogen atom is 13.6 eV.
An excited state of H atom emits a photon of wavelength lamda and returns in the ground state. The principal quantum number of excited state is given by:
The energy levels in a certain single electron species are all 800% higher in magnitude than corresponding levels of atomic hydrogen. A cartain transition of the electron from the n^(th) excited state to the next higher level is possible with a photonn of wavelength 72 nm . Find the value of n . Given : ((1)/(R) = 90 nm)
In a hydrogen atom, the electron is in n^(th) excited state. It may come down to second excited state by emitting ten different wavelengths. What is the value of n ?
Find the quantum number n corresponding to the excited state of He^(+) ion if on transition to the ground state that ion emits two photons in succession with wavelengths state that ion emits two photons in succession with wavelengths 1026.7 and 304Å. (R = 1.096 xx 10^(7)m^(_1)
