The ratio between total acceleration of the e | vedclass

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Question

(b) Acceleration $a \propto \frac{{{v^2}}}{r}$
where $v \propto \frac{Z}{n}$ and $r \propto \frac{{{n^2}}}{Z}$ ==> $a \propto \frac{{{Z^3}}}{{{n^4

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(b) Acceleration $a \propto \frac{{{v^2}}}{r}$
where $v \propto \frac{Z}{n}$ and $r \propto \frac{{{n^2}}}{Z}$ ==> $a \propto \frac{{{Z^3}}}{{{n^4}}}$
Since both are in ground state i.e., $n = 1$
so $a \propto Z^3$ ==> $\frac{{{a_{H{e^ + }}}}}{{{a_H}}} = {\left( {\frac{{{Z_{H{e^ + }}}}}{{{Z_H}}}} \right)^3} = {\left( {\frac{2}{1}} \right)^3} = \frac{8}{1}$.

The ratio between total acceleration of the electron in singly ionized helium atom and hydrogen atom (both in ground state) is

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