In a hydrogen atom,an electron is transferred | vedclass

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(D) The radius of an orbit in a hydrogen atom is given by $r_n = 0.0529 	imes n^2 \ nm$. For $r_1 = 1.3225 \ nm$,$n_1^2 = 1.3225 / 0.0529 = 25$,so $n

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$0.4578 	imes 10^{-18}$

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(D) The radius of an orbit in a hydrogen atom is given by $r_n = 0.0529 	imes n^2 \ nm$. For $r_1 = 1.3225 \ nm$,$n_1^2 = 1.3225 / 0.0529 = 25$,so $n_1 = 5$. For $r_2 = 0.2116 \ nm$,$n_2^2 = 0.2116 / 0.0529 = 4$,so $n_2 = 2$. The energy of an electron in the $n$-th orbit is $E_n = -2.18 	imes 10^{-18} / n^2 \ J$. The energy change is $\Delta E = E_{n_2} - E_{n_1} = -2.18 	imes 10^{-18} 	imes (1/n_2^2 - 1/n_1^2)$. $\Delta E = -2.18 	imes 10^{-18} 	imes (1/4 - 1/25) = -2.18 	imes 10^{-18} 	imes (0.25 - 0.04) = -2.18 	imes 10^{-18} 	imes 0.21 = -0.4578 	imes 10^{-18} \ J$. The energy of the emitted radiation is $|\Delta E| = 0.4578 	imes 10^{-18} \ J$.

In a hydrogen atom,an electron is transferred from an orbit of radius $1.3225 \ nm$ to another orbit of radius $0.2116 \ nm$. What is the energy (in $J$) of the emitted radiation?

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