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 $\frac{dP}{dr}=\frac{4Z^3}{a_0^3}(2r-\frac{2Z{r}^2}{a_0})e^{\frac{-27r}{a_0}};$
Then which of the following statement is / are correct?

• A. At the point of maximum value of radial distribution function $\frac{dP\left(r\right)}{dr}=0$
• B. Most probable radius of $L{i}^{2+}$ is $\frac{a_0}{3}\text{pm}$
  
• 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)

Most probable radius of hydrogen atom is $a_0\text{pm}$

$\left(a\right)$ At the point of maximum value of radial distribution function
$\frac{dP}{dr}=0\left(b\right)(2r-\frac{2Z{r}^2}{a_0})=0;r=\frac{a_0}{Z}$
Where $Z=3$ for $L{i}^{2+}$
Hence, most probable radius for $L{i}^{2+}=\frac{a_0}{3}\text{pm}$
$\left(c\right)$ $Z=2$ for the $H{e}^{+}$
Hence, most probable radius for $H{e}^{+}=\frac{a_0}{2}\text{pm}$
(d) For hydrogen, $Z=1$
Hence, most probable radius for hydrogen $=\frac{a_0}{1}=a_0\text{pm}$