Question

If 4 times the most probable distance of electron of a 1s orbital in a unielectronic atom /ion is given by $0.529A^{\circ }$ then the atomic number of the unielectronic species will be. [Given: Wave function of 1s orbital, $\Psi =\sqrt{\left(\frac{Z^3}{\pi a_0^3}\right)}{e}^{-Zr/a_0}$ where $a_0$ = radius of first Bohr's orbital in H - atom = (52.9 pm) and Z is atomic number of unielectronic species]

Answer 1

Given,
$4\times r_{mp}=0.529A^{\circ }$ .......(1)
For most probable distance of electron
$\frac{d}{dr}\left(4\pi r^2{\Psi }^2\right)=0$

$\Rightarrow \frac{d}{dr}\left(4\pi r^2\frac{Z^3}{\pi a_0^3}{e}^{\frac{-2Zr}{a_0}}\right)=0$
$\Rightarrow \frac{d}{dr}\left(r^2{e}^{\frac{-2Zr}{a_0}}\right)=0$
$\Rightarrow e^{\frac{-2Zr}{a_0}}[\frac{-2Z}{a_0}{r}^2+2r]=0$

$\Rightarrow 2-\frac{2rZ}{a_0}=0$

Hence, $r_{mp}=\frac{a_0}{Z}=\frac{52.9\times {10}^{-2}}{Z}{A}^{\circ }.......\left(2\right)$
Given,
$4\times r_{mp}=0.529A^{\circ }$

$\Rightarrow 4\times \frac{0.529}{Z}=0.529$
$\Rightarrow Z=4$