An excited `He^(+)` ion emits two photons in succession with wavelengths 108.5nm 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 cm"))`
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"))
He^(+) is in n^(th) state. It emits two successive photons of wavelength 103.7nm and 30.7nm , to come to ground state the value of n is:
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:
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 :
A 100 eV electron collides with a stationary helium ion (He^(+)) in its ground state and excites to a higher level. After the collision , He^(+) ion emits two photons in succession with wavelength 1085 Å and 304 Å . Find the principal quantum number of the excite in its ground state and. Also calculate energy of the electron after the collision. Given h = 6.63 xx 10^(-34) J s .
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:
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
