InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
A hydrogen atom in a state of binding energy` 0.55eV` make a trasisition to a state of excitation energy of `10.2 eV` (i) what is the initial state of hydrogen atom? (ii) what is the final state of hydrogen atom? (iii) what is the wavelength of the photon emitted? |
|
Answer» (i) Let `n_(1)` be initial state of electron. Then ` E_(1)=-(13.6)/(n_(1)^(2))eV`, Here `E_(1) =- 0.85 eV`, therefore` -0.85=-(13.6)/(n_(1)^(2))` or `n_(1)=4` (ii) Let `n_(2)` be the final excition state of the electron. Since excition energy is a always measured with respect to the ground state, therefore ` DeltaE =13.6[1-1/n_(2)^(2) ]` here `DeltaE = 10.2 eV`, therefore, `10.2 = 13.6 [1-(1)/(n_(2)^(2))]` or `n_(2)=2` Thus , the electron jump from `n_(1) =4 to n_(2) =2` . (iii) the wavelength of the photon emitted for a transition between `n_(1) =4 n_(2) =2`, given by ` 1/lambda=R_(oo)[1/n_(2)^(2)-1/n_(1)^2] (or) 1/lambda=1.09xx10^(7)[1/2^(2)-1/4^(2)]= `4860 Å` |
|