In hydrogen atom, energy of first excited state is `- 3.4 eV`. Then, `KE` of the same orbit of hydrogen atom is.

Options

A. `+ 3.4 eV`
B. `+6.8 eV`
C. `-13.6 eV`
D. `+13.6 eV`

Answer

A

Explanation

(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`