A wave function for hydrogen atom is given below, where a_0 is the Bohr radius, "52.9177 pm". If the Radial node in 2s be at r_0. Then the value of r_0 in terms of a_0 is?
Here is the equation:
ψ_(2s) = color(red)(sqrt(2)/(8pi)) × [1/a_0]^"3/2" [2 - r/a_0]e^(-r//color(red)(a_0))
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r_0 = 1/2a_0
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r_0 = 2a_0
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r_0 = 4a_0
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r_0 = 2/sqrt(2)a_0
Here is the equation:
-
r_0 = 1/2a_0 -
r_0 = 2a_0 -
r_0 = 4a_0 -
r_0 = 2/sqrt(2)a_0
1 Answer
r_0 = 2a_0
http://hyperphysics.phy-astr.gsu.edu/
First of all, the wave function you have is not correct. The correct wave function is:
psi_(2s) = color(orange)(1/(4sqrt(2pi))) (1/a_0)^(3//2)(2 - r/a_0)e^(-r//color(orange)(2a_0))
The radial node occurs where the radial component
All the nonzero constants can be divided out to get:
0 = |[(2 - r/a_0)e^(-r//2a_0)]|_(r = r_0)
Since
0 = 2 - r_0/a_0
Therefore, the distance away from the nucleus at which there is a radial node is:
color(blue)(r_0 = 2a_0)
or about