The radius of a hydrogen atom in its ground s | vedclass

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(B) The radius of an orbit in a hydrogen atom is given by $r_n = r_0 n^2$,where $r_0$ is the radius of the ground state $(n=1)$.Therefore,the ratio of

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$2$

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(B) The radius of an orbit in a hydrogen atom is given by $r_n = r_0 n^2$,where $r_0$ is the radius of the ground state $(n=1)$.Therefore,the ratio of the final radius $r_f$ to the initial radius $r_i$ is given by $\frac{r_f}{r_i} = \left( \frac{n_f}{n_i} ight)^2$.Given $r_i = 5.3 	imes 10^{-11} \ m$ at $n_i = 1$ and $r_f = 21.2 	imes 10^{-11} \ m$ at $n_f = n$.Substituting the values: $\frac{21.2 	imes 10^{-11}}{5.3 	imes 10^{-11}} = \left( \frac{n}{1} ight)^2$.$4 = n^2$.Taking the square root,we get $n = 2$.

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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