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(i) Which is highest frequency photon that can be emitted from hydrogen atom? What is wavelength of this photon? (ii) Calclate the longest wavelength transition in the Pashen series of He^(+). (iii) Calculate the ratio of the wavelength of first and the ultimate line of Balmer sereis of Li^(2+)? |
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Answer» Solution :(i) Highest frequency photon is emitted when electron comes from INFINITY to 1ST energy level. `E=-(13.6z^(2))/(1^(2))=-13.6eV` or `13.6xx1.6xx10^(-19)` Joule `=2.76xx10^(-18)` Joule `E=hv` `:.v=E/h=(2.176xx10^(-18)J)/(6.626xx10^(-34)Js)=0.328xx10^(18)Hz` `v=c/(LAMDA):.lamda=(3xx10^(80))/(0.328xx10^(16)Js)=9.146xx10^(-8)m` (iii) `barv=R_(H)xxZ^(2)[1/(n_(1)^(2))-1/(n_(2)^(2))]` For He Z=2, For Paschen series `n_(1)=3` For longest wavelength `n_(2)=4` `1/(lamda)=109678xx(2)^(2)xx[1/(3^(2))-1/(4^(2))]=109678xx4xx[1/9-1/16]=109678xx4xx7/144` `lamda=4689Å` (iii) WAVE number of first line of Balmer, `barv_(1)=RZ^(2)[1/(2^(2))-1/(3^(2))]=(5xx9r)/36=(5R)/4` Wavelength of first line of Balmer `=4/(5R)` Wave number of ultimate line ofBalmer `barv_(2)=RZ^(2)[1/(2^(2))-1/(OO)]=(9R)/4` `:.` Wavelength of ultimate of Balmer `=4/(9R)` `:.` Ratio `=9/5` |
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