What is the radius of the ground state of He⁺? A) 0.0529 nm B) 0.193 nm C) 0.0265 nm D) 0.106 nm E) 0.212 nm

Question

Quizplus · Accepted Answer

In order to find the radius of the ground state, we need to use the formula: $r=\frac{n^2h^2}{4\pi^2mZe^2}$ where $n$ is the principal quantum number, $h$ is Planck's constant, $m$ is the reduced mass of the system, $Z$ is the atomic number, and $e$ is the elementary charge. For He^{, we have:

$n=1$
$m=\frac{m_{He}m_{S}}{m_{He}+m_{S}}$
$Z=2+16+1=19$

Substituting these values into the formula, we get:

$r_{He^{}=\frac{(1)^2(6.626 imes10^{-34}\ J\cdot s)^2}{4\pi^2((4 imes10^{-26}\ kg)(19)(1.602 imes10^{-19}\ C))^2} = 0.0265\ nm$

For}, we have:

$n=1$
$m=\frac{m_{S}m_{S}}{m_{S}+m_{S}}$
$Z=16+16=32$

Substituting these values into the formula, we get:

$r_{}}=\frac{(1)^2(6.626 imes10^{-34}\ J\cdot s)^2}{4\pi^2((2 imes10^{-26}\ kg)(32)(1.602 imes10^{-19}\ C))^2} = 0.0529\ nm$ Therefore, the radius of the ground state of He⁺ is: $r=r_{He^}+r_{}=0.0265\ nm+0.0529\ nm=0.0794\ nm\approx0.08\ nm$ The closest answer choice to this value is C) 0.0265 nm.

What Is the Radius of the Ground State of He+

What Is the Radius of the Ground State of He⁺