Question

The Schrodinger wave equation for hydrogen atom is
${\Psi }_{2s}=\frac{1}{4\sqrt{2\pi }}{\left(\frac{1}{a_0}\right)}^{3/2}\left(2-\frac{r_0}{a_0}\right)e^{-r_0/a_0}$
Where $a_0$ is Bohr's radius. If the radial node in $2s$ be at $r_0$, then find $r_0$ in terms of $a_0$.

• A. $r_0=a_0$
• B. $r_0=4a_0$
• C. $r_0=2a_0$
• D. $r_0=7a_0$

Answer 1

Answer: (C)

The probability of finding of $2s$ electron at a point,
${\Psi }_{2s}^2=\frac{1}{32\pi }{\left(\frac{1}{a_0}\right)}^3{\left(2-\frac{r}{a_0}\right)}^2{e}^{-\frac{2r}{a_0}}$
Node is the point at which probability of finding an electron is zero.
It means the value of ${\Psi }_{2s}^2$ is zero when $r=r_0$.
$\frac{1}{32\pi }{\left(\frac{1}{a_0}\right)}^3{\left(2-\frac{r_0}{a_0}\right)}^2{e}^{-\frac{2r}{a_0}}=0$
$\Rightarrow 2-\frac{r_0}{a_0}=0$
$\therefore r_0=2a_0$