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(A) The energy of an electron in the $n$-th state of a hydrogen atom is given by $E_n = -\frac{|E|}{n^2}$.Given,the initial state is $n_A = 3$,so the

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$6$

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(A) The energy of an electron in the $n$-th state of a hydrogen atom is given by $E_n = -\frac{|E|}{n^2}$.Given,the initial state is $n_A = 3$,so the initial energy is $E_{n_A} = -\frac{|E|}{3^2} = -\frac{|E|}{9}$.When a photon of energy $\Delta E = \frac{|E|}{12}$ is absorbed,the electron transitions to state $n_B$ with energy $E_{n_B} = -\frac{|E|}{n_B^2}$.The energy conservation equation is $E_{n_B} - E_{n_A} = \Delta E$.Substituting the values: $-\frac{|E|}{n_B^2} - (-\frac{|E|}{9}) = \frac{|E|}{12}$.Dividing by $|E|$: $-\frac{1}{n_B^2} + \frac{1}{9} = \frac{1}{12}$.Rearranging the terms: $\frac{1}{n_B^2} = \frac{1}{9} - \frac{1}{12}$.Finding a common denominator: $\frac{1}{n_B^2} = \frac{4 - 3}{36} = \frac{1}{36}$.Therefore,$n_B^2 = 36$,which gives $n_B = 6$.

Similar Questions

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