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An electronic transition in hydrogen atom result in the formation of `Halpha` line of Hydrogen in Lyman series, the energies associated with the electron in each of the orbits involved in the transition (in `kcal mol^(-1)`) areA. `-313.6-34.84`B. `-313.6-78.4`C. `-78.4-34.84`D. `-78.4-19.6` |
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Answer» Correct Answer - B Energy of an electron in bth orbit `E_(n)=(2pi^(2)Z^(2)e^(4))/(n^(2)h^(2))` On substituting the values of k m e and h we get `E_(n)=(2.172xx10^(-18)Z^(2))/n^(2)J atom^(-1)` `=(1311.8Z^(2))/n^(2)kJ mol^(-1)=-(313.52Z^(2))/n^(2)kcal mol^(-1)` [.: 1 kacal =4.184kJ] For H - atom Z=1 For Lyman series `n_(1)=1 n_(2)=2` `=-(313.52xx (1)^(2))/(1)^(2)kcalmol^(-1)` `=-313.52kcal mol^(-1)` `=-313.6 kcal mol^(-1)` Energy of electron in `n_(2)` orbit `=-(313.52xx(1)^(2))/(2)^(2)kcalmol^(-1)` `-(313.52)/4 kcal mol^(-1)` `=-78.38 kcal mol^(-1)` |
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