The radius of the orbit of an electron in a H | vedclass

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(B) The radius of an orbit in a hydrogen atom is given by $R_n = n^2 a_0$,where $a_0 = 0.53 \mathring{A}$.For the initial state: $2.12 = n_1^2 	imes

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Absorbed a photon of energy $1.89 \ eV$

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(B) The radius of an orbit in a hydrogen atom is given by $R_n = n^2 a_0$,where $a_0 = 0.53 \mathring{A}$.For the initial state: $2.12 = n_1^2 	imes 0.53 \implies n_1^2 = 4 \implies n_1 = 2$.For the final state: $4.77 = n_2^2 	imes 0.53 \implies n_2^2 = 9 \implies n_2 = 3$.Since the radius increases,the electron has transitioned from a lower energy level to a higher energy level,meaning the atom has absorbed a photon.The energy of the absorbed photon is $\Delta E = E_3 - E_2 = -13.6 \left( \frac{1}{3^2} - \frac{1}{2^2} ight) \ eV$.$\Delta E = -13.6 \left( \frac{1}{9} - \frac{1}{4} ight) = -13.6 \left( \frac{4-9}{36} ight) = -13.6 \left( -\frac{5}{36} ight) \ eV$.$\Delta E = \frac{68}{36} \ eV \approx 1.89 \ eV$.

The radius of the orbit of an electron in a $H$-atom changes from $2.12 \mathring{A}$ to $4.77 \mathring{A}$. The $H$-atom has:

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