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The Schrodinger wave equation for hydrog...

The Schrodinger wave equation for hydrogen atom is `Psi_(2s) = (1)/(4sqrt(2pi)) ((1)/(a_(0)))^(3//2) (2 - (r)/(a_(0))) e^(-r//a_(0))` , where `a_(0)` is Bohr's radius . If the radial node in 2s be at `r_(0)` , then `r_(0)` would be equal to :

Text Solution

Verified by Experts

Given, `Phi_(2s)=(1)/(4sqrt(2)pi)((1)/(a_(0)))^(3//2)[2-(r_(0))/(a_(0))]e^(-r//a_(0))`
`Phi_(2s)^(2)=0` at node
`therefore2-(r_(0))/(a_(0))=0" or "r_(0)=2a_(0)`.
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Knowledge Check

  • The Schrodinger wave equation for hydrogen atom is psi_(2s) =1/(4sqrt(2pi)) (1/(a_(0)))^(3//2)(2-r/(a_(0)))e^(-t//a_(0)) where a_0 is Bohr's radius. If the radial node in 2 s be at r_0 , then r_0 would be equal to

    A
    `(a_(0))/2`
    B
    `2a_(0)`
    C
    `sqrt(2)a_(0)`
    D
    `(a_(0))/(sqrt(2))`

  • The schrodinger wave equation for hydrogen atom is Psi_(2s)=1/(4sqrt2pi)(1/a_0)^(1//2) [2-r_0/a_0]e^(-2//a_0) where a_0 is Bohr radius. If the radial node in 2s be at r_0 , then find r in terms of a_0

    A
    `r_0=2a_0`
    B
    `2r_0=a_0`
    C
    `3//2r_0=a_0`
    D
    `r_0=a_0`

  • The Schrodinger wave equation for hydrogen atom of 4s- orbital is given by : Psi (r) = (1)/(16sqrt4)((1)/(a_(0)))^(3//2)[(sigma^(2) - 1)(sigma^(2) - 8 sigma + 12)]e^(-sigma//2) where a_(0) = 1^(st) Bohr radius and sigma = (2r)/(a_(0)) . The distance from the nucleus where there will be no radial node will be :

    A
    `r = (a_(0))/(2)`
    B
    `r = 3a_(0)`
    C
    `r = a_(0)`
    D
    `r = 2a_(0)`