What are the frequency of a photon emitted during a transition from n = 5 state to the n = 2 state in the hydrogen atom?

Calculate the energy of radiation emitted for the electronic transition from infinity to ground state for hydrogen atom. Given c = 3xx10^(8)" ms"^(-1), R_(H) = 1.09678 xx 10^(7) m^(-1), h = 6.6256 xx 10^(-34)" J s"^(-1)

Radiation of certain frequency is emitted when electronic transition occurs from n = 4 to n = 2 in He^(+) which transition in H atom gives radiation of the same frequency?

For one-electron species, the wave number of radiation emitted during the transition of electron from a higher energy state (n_(2)) to a lower energy state (n_(1)) is given by: bar v =(1)/(lamda)=R_(H) xx Z^() ((1)/(n_(1)^(2)) (1)/(n_(2)^(2))) where R_(H)=(2 pi m_(s) k^(2) c^(4))/(h^(3) c) is Rydberg constant for hydrogen atom. Now, considering nuclear motion, the accurate measurement would be obtained by replacing mass of electron (m_(e)) by the reduced mass (mu) in the above expression, defined as mu =(m_(n)xx m_(e))/(m_(n) +m_(e)) where m_(n) = mass of nucleus. For Lyman series, n_(t) =1 (fixed for all the lines) while n_(2) = 2, 3, 4 .... For Balmer series: n_(1) = 2 (fixed for all the lines) while n_(2) = 3,4,5 .... The ratio of the wave numbers for the highest energy transition of electron in Lyman and Balmer series of hydrogen atom is