The ratio of the centripetal accelerations of | vedclass

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(C) The centripetal acceleration of an electron in the $n^{th}$ orbit is given by $a_c = \frac{v^2}{r}$. Since $v \propto \frac{1}{n}$ and $r \propto

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$\frac{h}{2 \pi}$

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(C) The centripetal acceleration of an electron in the $n^{th}$ orbit is given by $a_c = \frac{v^2}{r}$. Since $v \propto \frac{1}{n}$ and $r \propto n^2$,we have $a_c \propto \frac{(1/n)^2}{n^2} = \frac{1}{n^4}$.Given the ratio of accelerations for two successive orbits $n$ and $(n-1)$ is $\frac{a_n}{a_{n-1}} = \frac{(n-1)^4}{n^4} = \frac{16}{81}$.Taking the fourth root,we get $\frac{n-1}{n} = \frac{2}{3}$,which implies $3n - 3 = 2n$,so $n = 3$.The two orbits are $n = 3$ and $n = 2$.The angular momentum of an electron in the $n^{th}$ orbit is $L_n = \frac{nh}{2\pi}$.The change in angular momentum is $\Delta L = L_n - L_{n-1} = \frac{nh}{2\pi} - \frac{(n-1)h}{2\pi} = \frac{h}{2\pi}(n - n + 1) = \frac{h}{2\pi}$.

The ratio of the centripetal accelerations of the electron in two successive orbits of hydrogen is $81: 16$. Due to a transition between these two states,the angular momentum of the electron changes by ($h$ - Planck's constant).

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