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

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(A) The radius of a hydrogen atom is given by the formula $r_n = a_0 n^2$,where $a_0 = 0.53 \ Å$ is the Bohr radius and $n$ is the principal quantum n

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$n=2$

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(A) The radius of a hydrogen atom is given by the formula $r_n = a_0 n^2$,where $a_0 = 0.53 \ Å$ is the Bohr radius and $n$ is the principal quantum number.Given,the ground state radius $r_1 = 0.53 \ Å$ (for $n_1 = 1$).The radius of the excited state is $r_2 = 2.12 \ Å$ (for $n_2 = n$).Using the relation $r \propto n^2$,we have:$\frac{r_2}{r_1} = \left(\frac{n_2}{n_1}\right)^2$Substituting the values:$\frac{2.12}{0.53} = \left(\frac{n}{1}\right)^2$$4 = n^2$$n = \sqrt{4} = 2$Therefore,the principal quantum number of the final state is $n=2$.

The radius of a hydrogen atom in the ground state is $0.53 \ Å$. After collision with an electron,it is found to have a radius of $2.12 \ Å$. The principal quantum number $n$ of the final state of the atom is:

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