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The energy level diagram of an element is given below. Identify, by doing necessary calculations,-085 eV which transition corresponds to the emission of a spectral line of wavelength 102.7 mm. |
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Answer» Solution :If the WAVELENGTH of the emitted SPECTRAL line be `lambda = 102.7 nm=102.7 xx 10^(-9) m`, the energy of emitted photon will be `E = (hc)/(lambda) = (6.626xx 10^(-34) xx 3xx 10^(-8))/(102.7 xx 10^(-9)J)` ` = (6.626xx 10^(-34) xx 3 xx 10^(8))/(102.7 xx 10^(-9)) J` ` = (6.626 xx 10^(-34) xx 3 xx 10^(8))/(102.7 xx 10^(-9) xx 1.602 xx 10^(-19)) eV = 12.08 eV` From the energylevel diagram , we findthata radiation photon of 102.7 nm can be emitted onlywhenan electron transitiontakes place from - 1.5 eV energyto- 13.6V energylevel becausethen . `E_(1) = E_(2) = - 1.5 - (-13.6) eV = + 12.1 eV` Hence, TRANSITION D REPRESENT emission of spectral line of wavelenght 102.7 nm . 
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