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Find the distance at which probability of finding electron is maximum for `1s` orbital in a He atom. The wave function orbital given as.
`psi_(1s)=(4)/(a_(0)^(3//2))e^((2r)/(a_(0))`

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Probability distribution function is `P(r )= psi^(2) = 4pir^(2)=((16)/(a_(0)^(3)))e^(((4r)/(a_(0))))4pir^(2)`
`implies P(r ) = k. r^(2).e.^(4r)/(a_(0))`
differentating `(dP(r ))/(dr) = 2r.e.^(4r)/(a_(0)) - ((4)/(a_(0)))r^(2).e^(4r)/(a_(0))=0`
`implies 1=(2r)/(a_(0)) implies r=(a_(0))/(2)`
implies probability of finding electron is maximum at distance `(a_(0))/(2)` from nucleus.
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