A hydrogen atom in a having a binding of `0.85eV`makes transition to a state with excited energy `10.2eV`(a) identify the quantum number n of theupper and the lower energy state involved in the transition (b) Find the wavelength of the emitted radiation

Explanation

a. Let `n_(1)` be the initial state of electron, Then.
`E_(1) = - (13.6)/(n_(1)^(2)) e V`
Here, `E_(1) = - 0.85 e V`
`:. - 0.85 = - (13.6)/(n_(1)^(2))`
or `n_(1) = 4`
b. Let `n_(2)` be the final excitation state of electron. Since excitation energy is always measured with respect to the ground state, therefore
`Delta E = 13.6 [1 - (1)/(n_(2)^(2))]`
here `Delta E = 10.2 e V`
`:. 10.2 = 13.6 [1 - (1)/(n_(2)^(2))]`
or `n_(2) = 2`
Thus the electron jumps from `n_(2) = 4` to `n_(2) = 2`
c. The wavelength of the photon emitted for a transition between `n_(2) = 4` to `n_(2) = 2` is given by
`(1)/(lambda) = R_(oo) [(1)/(n_(2)^(2)) - (1)/(n_(1)^(2))]`
`(1)/(lambda) = 1.097 xx 10^(7) [(1)/(2^(2)) - (1)/(4^(2))]`
`lambda = 4860 Å`