The Schrödinger wave equation for the hydrogen atom is:
\[\psi_{2s} = \frac{1}{4\sqrt{2\pi}} \left( \frac{1}{a_0} \right)^{3/2} \left( 2 - \frac{r_0}{a_0} \right) e^{-\frac{r_0}{2a_0}} \], where \(a_0\) is Bohr's radius. If the value of \(a_0\) is 53 pm, then what is the value of the distance of the radial node (\(r_0\) in pm) from the nucleus?

To find the distance of the radial node for the given wave function ψ 2 s ​ , we need to understand the relationship between the function and its radial nodes. A radial node occurs where the radial part of the wave function equals zero.
The given wave function for the hydrogen atom in the 2 s state is:
ψ 2 s ​ = 4 2 π ​ 1 ​ ( a 0 ​ 1 ​ ) 3/2 ( 2 − a 0 ​ r 0 ​ ​ ) e − 2 a 0 ​ r 0 ​ ​
To find the radial node, set the bracketed expression to zero because that's where the function changes sign. So, we solve:
2 − a 0 ​ r 0 ​ ​ = 0
Solving for r 0 ​ :
a 0 ​ r 0 ​ ​ = 2
r 0 ​ = 2 a 0 ​
We are given that the Bohr radius a 0 ​ is 53 pm. Substituting this into the equation:
r 0 ​ = 2 × 53 pm
r 0 ​ = 106 pm
So the distance of the radial node from the nucleus is 106 pm. This means there is a point at 106 pm from the nucleus where the probability density is zero, which is characteristic of a node in quantum mechanics.

Answered by RyanHarmon181 | 2025-07-06