Question

The radial distribution functions [P(r)] 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
$\frac{dP}{dr}=\frac{4Z^3}{a_0^3}(2r-\frac{2Z{r}^2}{a_0})e^{-2Zr/a_0}$

Then which of the following statements is/are correct?

• A. At the point of maximum, value of radial distribution function $\frac{dP\left(r\right)}{dr}=0;$ one antinode is present
• B. Most probable radius of $L{i}^{2+}is\frac{a_0}{3}pm$
• C. Most probable radius of $H{e}^{+}is\frac{a_0}{2}pm$
• D. Most probable radius of hydrogen atom is $a_{0}pm$

Answer 1

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

The correct options are
A At the point of maximum, value of radial distribution function $\frac{dP\left(r\right)}{dr}=0;$ one antinode is present
B Most probable radius of $L{i}^{2+}is\frac{a_0}{3}pm$
C Most probable radius of $H{e}^{+}is\frac{a_0}{2}pm$
D Most probable radius of hydrogen atom is $a_{0}pm$

At the point of maximum, the slope of radial distribution curve becomes zero. i.e,
$\frac{dP}{dr}=0$
this is only possible when :
$(2r-\frac{2Z{r}^2}{a_0})=0r=\frac{a_0}{Z}$
where Z = 3 for $L{i}^{2+}$ and Z = 2 for the $H{e}^{+}$; Z = 1 for hydrogen.
Therefore, a , b, c, d are correct.