Question

The wavefunction of a 2s electron is given by ${\psi }_{2s}=\frac{1}{4\sqrt{2}\pi }\left(\frac{1}{a_0}{)}^{\frac{3}{2}}\right(2-\frac{r}{a_0})e^{\frac{-r}{a_0}}$
It has a node at $r=r_0$. The value of $\frac{r_0}{a_0}$ is___

Answer 1

${\psi }^2=\frac{1}{32{\pi }^2}\left(\frac{1}{a_0}{)}^3\right(2-\frac{r}{a_0}{)}^2{e}^{\frac{-2r}{a_0}}$
When $r=r_0$, it has a node (i.e.) ${\psi }^2=0$
$\frac{1}{32{\pi }^2}\left(\frac{1}{a_0}{)}^3\right(2-\frac{r_0}{a_0}{)}^2{e}^{\frac{-2r_0}{a_0}}=0$
From this relation it is clear that any of these terms
$(i.e)\frac{1}{32{\pi }^2}\left(or\right)\left(\frac{1}{a_0}{)}^3\right(or\left)\right(2-\frac{r_0}{a_0}{)}^2\left(or\right)(e^{\frac{-2r_0}{a_0}}$) has to be equal to zero.
It is only possible for $(2-\frac{r_0}{a_0})$ to be equal to 0, all the others cannot ever become zero

Hence, $2-\frac{r_0}{a_0}=0\left(or\right)\frac{r_0}{a_0}=2$