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A difference of 2.3 eV separates two ene...

A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom make a transition from the upper level to the lower level?

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`5.6 xx 10^(14) Hz`
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For the energy levels in an atom, which one of the following statement(s) is (are) correct ?

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