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(C) The area of an orbit is given by $A = \pi r^2$. Since $r \propto n^2$,we have $A \propto (n^2)^2 = n^4$.Given $\frac{A_1}{A_2} = \frac{4}{1}$,then

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$1 : 2\sqrt{2}$

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(C) The area of an orbit is given by $A = \pi r^2$. Since $r \propto n^2$,we have $A \propto (n^2)^2 = n^4$.Given $\frac{A_1}{A_2} = \frac{4}{1}$,then $(\frac{n_1}{n_2})^4 = 4$,which implies $\frac{n_1}{n_2} = (4)^{1/4} = (2^2)^{1/4} = 2^{1/2} = \sqrt{2}$.The frequency of an electron in an orbit is given by $f \propto \frac{1}{n^3}$.Therefore,$\frac{f_1}{f_2} = (\frac{n_2}{n_1})^3 = (\frac{1}{\sqrt{2}})^3 = \frac{1}{2\sqrt{2}}$.Thus,the ratio is $1 : 2\sqrt{2}$.

If the ratio of the area of two orbits of an $H$ atom is $4 : 1$,then the ratio of the frequency of the $e^-$ in these two orbits is:

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