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(C) The radius of the $n^{	ext{th}}$ orbit of a hydrogen atom is given by $r_n = a_0 n^2$,where $a_0 = 0.529 \mathring{A}$.Given $r_n = 8.48 \mathrin

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$16$

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(C) The radius of the $n^{	ext{th}}$ orbit of a hydrogen atom is given by $r_n = a_0 n^2$,where $a_0 = 0.529 \mathring{A}$.Given $r_n = 8.48 \mathring{A}$,we have $8.48 = 0.529 	imes n^2$.$n^2 = \frac{8.48}{0.529} \approx 16$.Thus,$n = 4$.The energy of an electron in the $n^{	ext{th}}$ orbit is given by $E_n = \frac{E}{n^2}$,where $E$ is the ground state energy $(-13.6 	ext{ eV})$.Substituting $n = 4$,we get $E_4 = \frac{E}{4^2} = \frac{E}{16}$.Comparing this with $E/x$,we find $x = 16$.

The radius of a certain orbit of a hydrogen atom is $8.48 \mathring{A}$. If the energy of the electron in this orbit is $E/x$,then $x = . . . .$
(Given $a_0 = 0.529 \mathring{A}$,$E =$ energy of the electron in the ground state)

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The radius of a certain orbit of a hydrogen atom is $8.48 \mathring{A}$. If the energy of the electron in this orbit is $E/x$,then $x = . . . .$(Given $a_0 = 0.529 \mathring{A}$,$E =$ energy of the electron in the ground state)