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For 3s orbital of hydrogen atom, the nor...

For `3s` orbital of hydrogen atom, the normalised wave function is
`Psi_(3s)=(1)/((81)sqrt(3pi))((1)/(a_(o)))^(3//2)[27-(18r)/(a_(o))+(2r^(2))/(a_(o)^(2))]e^((-r)/(3a_(o)))`
If distance between the radial nodes is d, calculate rthe value of `(d)/(1.73a_(o))`

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At nodal point `v = 0` from the given wave function we find that `psi = 0 `at following values of r
`[27 - 18 ((r )/(a_(0))) + 2 ((r )/(a_(0)))^(3)] = 0 `
Solving `r_(0)//a_(p)` we get
`(r_(0))/(a_(0)) = (18 +- sqrt(10^(2) - 216))/(4) = (16 +- 10.4)/(2)`
Hence `s = 14.2a_(0) `and `r_(0) = 3 9a_(0)`
Besides there is a node at `r = oo`
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CENGAGE CHEMISTRY ENGLISH-ATOMIC STRUCTURE-Concept Applicationexercise(4.3)
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