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In hydrogen atom, energy of first excited state is `- 3.4 eV`. Then, `KE` of the same orbit of hydrogen atom is.A. ` + 3.4 eV`B. `+ 6.8 eV`C. `-13.6 eV`D. `+ 13.6 eV` |
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Answer» Correct Answer - A (a) `because` Total energy `(E_n) = KE + PE` In the first excited state =`(1)/(2) mv^2 + [-(Ze^2)/( r)]` =`+ (1)/(2) (Ze^2)/( r) - (Ze^2)/( r)` `- 3.4 e V = - (1)/(2) (Ze^2)/( r)` `:. KE = (1)/(2) (Ze^2)/(2) = + 3.4 eV`. |
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