The radial probability distribution function for a $3s$ orbital has two nodes. Calculate the distance of the outer node from the nucleus in terms of the Bohr radius $a_0$.

Solution

1. The radial wave function for $3s$ is $R_{30}(r)\propto (6-6\rho +{\rho }^2)e^{-\rho /2}$ where $\rho =\frac{2Zr}{3a_0}$.
2. Nodes occur where $6-6\rho +{\rho }^2=0$.
3. Solving the quadratic equation ${\rho }^2-6\rho +6=0$ gives $\rho =3\pm \sqrt{3}$.
4. The outer node is at $\rho =3+\sqrt{3}\approx 4.732$.
5. Substituting $\rho =\frac{2r}{3a_0}$ for hydrogen ($Z=1$), $r=\frac{4.732\times 3}{2}{a}_0\approx 7.098a_0$.
6. Hence the answer is (A).