When an electron makes a transition from (n + 1) state to `n^(th)` state, the frequency of emitted radiations is related to ‘n’ according to `( n gt gt 1)`

The electron in a hydrogen atom makes a transition from n = n_(1) to n = n_(2) state. The time period of the electron in the initial state is eight times that in the electron in the initial state is eight times that is the final state. The possible values of n_(1) and n_(2) are

When a hydrogen atom emits a photon during a transition from n = 4 to n = 2, its recoil speed is about ,

Why is the transition from n=3 to n=1 gives a spectral triplet

According to the Bohr theory for the atomic spectrum of hydrogen the energy levels of the proton -electron system depends, on the quantum number n. In an electron transition from a higher quantum level , n_(2) to the lower level n_(1) , radiation is emitted. The frequency, v of the radiation is given by h v = (E_(n_(2)) - E_(n_(1))) where, h is Planck.s constant and E_(n_(2)), E_(n_(1)) are the energy level values for the quantum number n_(2) and n_(1) . A useful formula for the wavelength, lambda = c//v is given by = lambda(Å) = (912)/(Z^(2)) xx ((n_(2)^(2)n_(1)^(2))/(n_(2)^(2)-n_(1)^(2))) where, Z = atomic number of any one electron species viz H, He^(+), Li^(2+), Be^(3+) ,..... For hydrogen Z=1 In the above formula, when n_(2) rarr oo , we have lambda = (912)/(1^(2)) xx n_(1)^(2) . This values of lambda is known as the series limit for the given value of n_(1) . Calculate the value of n_(1) for the series limit lambda = 8208 Å

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

When an electron jumps from a level n = 4 to n = 1 , the momentum of the recoiled hydrogen atom will be