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The kinetic energy of an electron in H...

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 report the atom.

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K.E. `=6.04` in 3rd orbit.
`E_("total")=K.E.+P.E.`
`=K.E.=2xxK.E.`
`implies-K.E.=-6.04eV`
`E_(1)` for `H=-13.6eV` and not for any orbit `E=-6.04eV` for H atom. Thus, atom for which K.E. is given is other than H
`E_(n)` H like atom `=E_(nH)=xxZ^(2)`
`(E_(1))/(n^(2))xxz^(2)implies6.04=13.6/(3^(2))xxz^(2)`
`Z^(2)=3.99~~4impliesZ=2`
`:.` The atom is `He^(+)impliesr_(n)=0.529xx(n^(2))/Z=0.526xx(36^(2))/2=2.3805Å`
Area `pirl^(2)=22/7xx(2.3805xx10^(-8))^(2)`
`=17.8xx10^(-16)cm^(2)`
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