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))

  1. r_0 = 1/2a_0

  2. r_0 = 2a_0

  3. r_0 = 4a_0

  4. r_0 = 2/sqrt(2)a_0

1 Answer
Nov 20, 2017

r_0 = 2a_0

http://hyperphysics.phy-astr.gsu.edu/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 R_(nl)(r) of the wave function goes to zero. But since there is no angular component Y_(l)^(m_l)(theta,phi) to a wave function for a spherical orbital (l = 0, m_l = 0), set psi_(2s) = 0.

All the nonzero constants can be divided out to get:

0 = |[(2 - r/a_0)e^(-r//2a_0)]|_(r = r_0)

Since e^(-r//2a_0) ne 0 for r in between 0 and oo (where nodes can occur), that can be divided out as well.

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 2 cdot "52.9177 pm" = "105.8354 pm".