The Bamber series for the hydrogen atom corresponds to electron that terminal in the state of quantum number `n = 2`.
a Find the longest wavelength photon emitted and determine its energy.
b Find the shortest wavelength photon emitted in the balmer series.

Explanation

The longest wevelength photon in the Balmer series result corresponding to transition from `n = 3 to n = 2`. Using the equation
`(1)/(lambda) = R ((1)/(n_(1)^(2)) - (1)/(n_(u)^(2))) implies (1)/(lambda)_(max) = R ((1)/(2^(2)) - (1)/3^(2)) = (5)/(36) R`
`implies lambda_(max) = (36)/(5 R) = (36)/(5 (1.097 xx 10^(7))) = 656.3 nm`
This lines wevelength lines in the red region of the visible spectrum.
The energy of emitted photon is
`E = (hc)/(lambda_(max)) = ((6.626 xx 10^(-34)) (3 xx 10^(8)))/((656.3 xx 10^(-9)))`
`= 3.03 xx 10^(-19) j = 1.89 e V`
b. The shoetest wevelength photon in the Balmer series is emitted when the electron makes a transition from `n = oo to n = 2`.
`(1)/(lambda_(max)) = R((1)/(2^(2)) - (1)/(oo))= (R)/(4)`
`implies lambda_(max) = (4)/(R) = (4)/(1.097 xx 10^(7)) = 364.6 nm`
This wavelength lines in the ultra-violet region`