The ratio of centripetal acceleration for an | vedclass

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(C) The centripetal acceleration $a_c$ of an electron in a Bohr orbit is given by $a_c = \frac{v^2}{r}$.For a hydrogen atom, the velocity $v$ in the $

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$625 : 81$

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(C) The centripetal acceleration $a_c$ of an electron in a Bohr orbit is given by $a_c = \frac{v^2}{r}$.For a hydrogen atom, the velocity $v$ in the $n^{th}$ orbit is proportional to $\frac{1}{n}$ $(v \propto \frac{1}{n})$ and the radius $r$ is proportional to $n^2$ $(r \propto n^2)$.Substituting these into the formula for centripetal acceleration:$a_c \propto \frac{(1/n)^2}{n^2} = \frac{1/n^2}{n^2} = \frac{1}{n^4}$.Therefore, the ratio of centripetal acceleration for the $3^{rd}$ and $5^{th}$ orbits is:$\frac{a_3}{a_5} = \frac{n_5^4}{n_3^4} = \left(\frac{5}{3}\right)^4 = \frac{625}{81}$.Thus, the ratio is $625 : 81$.

The ratio of centripetal acceleration for an electron revolving in $3^{rd}$ and $5^{th}$ Bohr orbit of hydrogen atom is

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