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Ratio of centripetal acceleration for an electron revolving in $3^{\text {rd }}$ orbit to $5^{\text {th }}$ Bohr orbit
of hydrogen atom is
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of hydrogen atom is
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The correct answer is:
$\frac{625}{81}$
Centripetal acceleration $\mathrm{a}=\frac{\mathrm{v}^{2}}{\mathrm{r}}$
$v \propto \frac{1}{n}$ and $r \alpha n^{2}$
$\therefore a=\frac{1}{n^{4}}$
$\therefore \frac{a_{5}}{a_{3}}=\frac{(3)^{4}}{(5)^{4}}=\frac{81}{125}$
$v \propto \frac{1}{n}$ and $r \alpha n^{2}$
$\therefore a=\frac{1}{n^{4}}$
$\therefore \frac{a_{5}}{a_{3}}=\frac{(3)^{4}}{(5)^{4}}=\frac{81}{125}$
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