The radius of stationary state (n=2) of hydro | vedclass

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(A) The radius of the $n^{th}$ orbit of a hydrogen-like species is given by the formula: $r_n = a_0 	imes \frac{n^2}{Z}$, where $a_0$ is the Bohr rad

Vedclass · Accepted Answer

$\frac{9}{8} x$

Vedclass · Answer

(A) The radius of the $n^{th}$ orbit of a hydrogen-like species is given by the formula: $r_n = a_0 	imes \frac{n^2}{Z}$, where $a_0$ is the Bohr radius, $n$ is the principal quantum number, and $Z$ is the atomic number.For the hydrogen atom $(Z=1)$ at $n=2$: $x = a_0 	imes \frac{2^2}{1} = 4a_0$, which implies $a_0 = \frac{x}{4}$.For the $He^{+}$ ion $(Z=2)$ at $n=3$: $r_3 = a_0 	imes \frac{3^2}{2} = a_0 	imes \frac{9}{2}$.Substituting $a_0 = \frac{x}{4}$ into the equation for $r_3$: $r_3 = (\frac{x}{4}) 	imes \frac{9}{2} = \frac{9}{8} x$.

The radius of stationary state $(n=2)$ of hydrogen atom is $x \ pm$. The radius of stationary state $(n=3)$ of $He^{+}$ ion (in $pm$) is

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