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(C) The centripetal acceleration $a_c$ 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 \pro

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

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(C) The centripetal acceleration $a_c$ 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 centripetal accelerations for two successive orbits $n$ and $n-1$ is $81:16$,we have $\frac{a_c(n-1)}{a_c(n)} = \frac{n^4}{(n-1)^4} = \frac{81}{16} = (\frac{3}{2})^4$.Thus,$n=3$ and $n-1=2$.The angular momentum of an electron in the $n$-th orbit is $L_n = \frac{nh}{2\pi}$.The change in angular momentum during the transition between these two states is $\Delta L = L_3 - L_2 = \frac{3h}{2\pi} - \frac{2h}{2\pi} = \frac{h}{2\pi}$.

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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