Question

The radial distribution functions $\left[P\right(r\left)\right]$ is used to determine the most probable radius, which is used to find the electron in a given orbital $\frac{dP\left(r\right)}{dr}$ for 1s-orbital of hydrogen like atom having atomic number $Z$, is $\frac{dP}{dr}=\frac{4Z^3}{a_0^3}(2r-\frac{2Z{r}^2}{a_0})e^{\frac{-2Zr}{a_0}}$. Which of the following statements is/are correct?

• A. At the maximum value of the radial distribution function $\frac{dP\left(r\right)}{dr}=0$, one antinode is present
• B. Most probable radius of $L{i}^{2+}$ is $\frac{a_0}{2}\text{pm}$
• C. Most probable radius of $H{e}^{+}$ is $\frac{a_0}{2}\text{pm}$
• D. Most probable radius of hydrogen atom is $a_0\text{pm}$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A At the maximum value of the radial distribution function $\frac{dP\left(r\right)}{dr}=0$, one antinode is present
B Most probable radius of $L{i}^{2+}$ is $\frac{a_0}{2}\text{pm}$
C Most probable radius of $H{e}^{+}$ is $\frac{a_0}{2}\text{pm}$
D Most probable radius of hydrogen atom is $a_0\text{pm}$

At the point of maximum value of radial distribution function(RDF),
$\frac{dP}{dr}=0$
$(2r-\frac{2Z{r}^2}{a_0}=0);r=\frac{a_0}{Z}$
where $Z=3$ for $L{i}^{2+}$ and $Z=2$ for $H{e}^{+}$; $Z=1$ for hydrogen.