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 Calculating n_(2) for the series limit = 22800 Å. Calculate lambda for the transition (n_(2) + 1) to (n_(1) - 1)

According to the Bohr theory for the atomic spectrum of hydrogen the e

1.	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 Calculating n_(2) for the series limit = 22800 Å. Calculate lambda for the transition (n_(2) + 1) to (n_(1) - 1)
Answer» <html><body><p>26266 Å<br/>22665 Å<br/>25266 Å<br/>22566 Å</p>Answer :A</body></html>