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:

Hydrogen atom from excited state comes to the ground state by emitting a photon of wavelength lambda . If R is the Rydberg constant, then the principal quatum number n of the excited state is

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

An excited hydrogen atom emits a photon of wavelength lambda in returning to the ground state. If 'R' is the Rydberg's constant, then the quantum number 'n' of the excited state is:

An excited He^(+) ion emits photon of wavelength lambda in returning to ground state from n^(th) orbit. If R is Rydberg's constant then :

An excited hydrogen atom returns to the ground state . The wavelength of emittted photon is lambda The principal quantium number fo the excited state will be :

A gas of hydrogen - like ion is perpendicular in such a way that ions are only in the ground state and the first excite state. A monochromatic light of wavelength 1216 Å is absorved by the ions. The ions are lifted to higher excited state and emit emit radiation of six wavelength , some higher and some lower than the incident wavelength. Find the principal quantum number of the excited state identify the nuclear charge on the ions . Calculate the values of the maximum and minimum wavelengths.

A photon of energy 12.09 eV is completely absorbed by a hydrogen atom initially in the ground state. The quantum number of excited state is

Hydrogen atoms are excited from ground state to the principal quantum number 5. Number of spectral lines observed will be

A gas of hydrogen like ions is prepared in a particular excited state A. if emit photons having wavelength equal to the wavelength of the first line of the lyman series together with photons of five other wavelength identify the gas and find the principal quantum number of the state A