Ratio of centripetal acceleration for an electron revolving in $3^{\text {rd }}$ orbit and $5^{\text {th }}$…
Ratio of centripetal acceleration for an electron revolving in $3^{\text {rd }}$ orbit and $5^{\text {th }}$ Bohr orbit of hydrogen atom is
$\frac{125}{81}$
$\frac{625}{81}$
$\frac{625}{27}$
$\frac{25}{9}$
Solution
Centripetal acceleration $a=\frac{v^2}{r}$
In Bohr orbit, the velocity $v \propto \frac{1}{n}$ and radius $r \propto n^2$ have functional dependencies on orbit number $n$.
Thus, the centripetal acceleration depends on the orbit number as follows: $a \propto \frac{1}{n^4}$
$\therefore \frac{a_3}{a_5}=\frac{5^4}{3^4}=\frac{625}{81}$