If the `P.E.` of an electron is `-68 eV` in hydrogen atom then find out `K.E.,E` of orbit where electron exist `&` radius of orbit.

Explanation

(i) `P.E=-2K.E`.
`-6.8=-2K.E.`
`(6.8)/(2) = K.E " "K.E. = 3.4 eV`
`(ii) " "E.=-K.E.`
`=-3.4 eV`
`(iii) " " "orbit" =2^(nd)`
`because E=-13.6xx(Z^(2))/(n^(2))`
`therefore 3.4 =- 13.6 xx(1^(2))/(n^(2))`
`implies n^(2)=(-13.6)/(-3.4) =4`
`i.e. n=2`
`(iv) r=0.529xx(n^(2))/(Z)Å`
`r=0.529xx((2)^(2))/(1)Å`
`= 0.529 xx 4Å = 2.16 Å`