Ratio of centripetal acceleration for an elec | vedclass

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(B) The centripetal acceleration is given by $a = \frac{v^2}{r}$.In Bohr's model,the velocity of an electron in the $n^{	ext{th}}$ orbit is $v \propt

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$\frac{625}{81}$

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(B) The centripetal acceleration is given by $a = \frac{v^2}{r}$.In Bohr's model,the velocity of an electron in the $n^{	ext{th}}$ orbit is $v \propto \frac{1}{n}$ and the radius is $r \propto n^2$.Substituting these proportionalities into the expression for acceleration,we get $a \propto \frac{(1/n)^2}{n^2} = \frac{1}{n^4}$.Therefore,the ratio of centripetal acceleration for the $3^{	ext{rd}}$ orbit $(n_1 = 3)$ and $5^{	ext{th}}$ orbit $(n_2 = 5)$ is $\frac{a_3}{a_5} = \frac{n_2^4}{n_1^4} = \frac{5^4}{3^4}$.Calculating the values,we get $\frac{625}{81}$.

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

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