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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) `:. ` Total energy `(E_(n))=KE +PE` In first excited state = `(1)/(2) mv^(2)+[-(Ze^(2))/(r)]` `=+(1)/(2) (Ze^(2))/(r)-(Ze^(2))/(r)` Energy of first excited state is 3.4 eV `-3.4 eV=-(1)/(2) (Ze^(2))/(r)` `:. KE=(1)/(2)(Ze^(2))/(2)` `=+3.4 eV` |
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