The kinetic energy of an electron in H like atom is `6.04 eV` Find the area of the third bohr orbit to which this electron belongs .Also repurt the atom

Explanation

`KE = 6.04` in third orbit
`E_(Total) = KE + PE = KE - 2 xx KE`
`rArr -KE = - 6.04 eV`
`E_(1) for H = 13.6 eV` and not for any orbit `E= 6.04 eV` for H atom .This atom for which KE is given is other than H
`E_(n)` of H-like atom = `E_(n) = Z^(2)`
`(E_(1))n^(2) xx Z^(2)rArr 6.04 = (13.6)/(3^(2)) xx Z^(2)`
`Z^(2) = 3.99 ~~ 4 rArr Z = 2`
Hence the atom to `He^(o+)` so
`rArr r_(n) = 0.529 xx (n^(2))/(Z) = 0.529 xx (3^(2))/(2) = 2.3805 Å`
Area `pi r^(2) = (22)/(7) xx (2.5805 xx 10^(-8))^(2)`
`= 17.8 xx 10^(-16)cm^(2)`