InterviewSolution
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InterviewSolution
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The wave function of 2s electron is given by `Psi_(2s) = (1)/(4 sqrt(2pi)) ((1)/(a_(0)))^(3//2) (2 - (r)/(a_(0))) e^(-r//a_(0))` It has a node at `r = r_(0)`. Find the relation between `r_(0) and a_(0)` |
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Answer» The probability of finding 2s electron will be : `Psi_(2s)^(2) = (1)/(32pi) ((1)/(a_(0)))^(3) (2- (r_(0))/(a_(0)))^(2) e^(-2r//a_(0))` Node is the point at which probability of finding electron is zero. Thus, `Psi_(2s)^(2) = 0` when `r = r_(0)` `:. (1)/(32pi) ((1)/(a_(0)))^(2) (2 - (r_(0))/(a_(0)))^(2) e^(-2r_(0)//a_(0)) = 0` In this expression, the only factor that can be zero is `(2 - (r_(0))/(a_(0)))` Thus, `2 - (r_(0))/(a_(0)) = 0 or r_(0) = 2 a_(0)` |
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