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(B) The radius of an orbit in a hydrogen atom is given by $r_n = a_0 n^2$,where $a_0$ is the Bohr radius $(5.3 	imes 10^{-11} \ m)$ and $n$ is the pr

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$n=2$

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(B) The radius of an orbit in a hydrogen atom is given by $r_n = a_0 n^2$,where $a_0$ is the Bohr radius $(5.3 	imes 10^{-11} \ m)$ and $n$ is the principal quantum number.Since $r \propto n^2$,we have the ratio:$\frac{r_f}{r_i} = \left(\frac{n_f}{n_i}ight)^2$Given $r_i = 5.3 	imes 10^{-11} \ m$ (for $n_i = 1$) and $r_f = 21.2 	imes 10^{-11} \ m$:$\frac{21.2 	imes 10^{-11}}{5.3 	imes 10^{-11}} = \left(\frac{n}{1}ight)^2$$4 = n^2$$n = 2$Therefore,the principal quantum number of the final state is $n=2$.

The radius of a hydrogen atom in its ground state is $5.3 \times 10^{-11} \ m$. After collision with an electron,it is found to have a radius of $21.2 \times 10^{-11} \ m$. What is the principal quantum number $n$ of the final state of the atom?

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