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(C) The orbital speed of an electron in the $n^{th}$ Bohr orbit is given by $v_n \propto \frac{1}{n}$.Given that the initial state is the ground state

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$9$

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(C) The orbital speed of an electron in the $n^{th}$ Bohr orbit is given by $v_n \propto \frac{1}{n}$.Given that the initial state is the ground state $(n_1 = 1)$ with speed $v_1$,and the final speed is $v_2 = \frac{v_1}{3}$.Using the relation $\frac{v_1}{v_2} = \frac{n_2}{n_1}$,we get $\frac{v_1}{v_1/3} = \frac{n_2}{1}$,which implies $n_2 = 3$.The radius of the $n^{th}$ orbit is given by $r_n \propto n^2$.Therefore,$\frac{r_2}{r_1} = \left(\frac{n_2}{n_1}\right)^2 = \left(\frac{3}{1}\right)^2 = 9$.Thus,the radius of the orbit in the excited state is $r_2 = 9 r_1$.

In a hydrogen atom in its ground state,the first Bohr orbit has radius $r_1$. The electron's orbital speed becomes one-third when the atom is raised to one of its excited states. The radius of the orbit in that excited state is (in $r_1$)

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