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(B) According to the Bohr model,the energy of an electron in the $n^{th}$ orbit is given by $E_n = -\frac{2.179 	imes 10^{-11}}{n^2} \ erg$.The chang

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$0.1911 	imes 10^{-10} \ erg$

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(B) According to the Bohr model,the energy of an electron in the $n^{th}$ orbit is given by $E_n = -\frac{2.179 	imes 10^{-11}}{n^2} \ erg$.The change in energy for a transition from $n_1 = 1$ to $n_2 = 3$ is $\Delta E = E_3 - E_1$.$\Delta E = -2.179 	imes 10^{-11} \left( \frac{1}{3^2} - \frac{1}{1^2} ight) \ erg$.$\Delta E = -2.179 	imes 10^{-11} \left( \frac{1}{9} - 1 ight) \ erg$.$\Delta E = -2.179 	imes 10^{-11} \left( -\frac{8}{9} ight) \ erg$.$\Delta E = 2.179 	imes 10^{-11} 	imes 0.8888 \ erg \approx 1.936 	imes 10^{-11} \ erg$.Converting to the given format: $1.936 	imes 10^{-11} \ erg = 0.1936 	imes 10^{-10} \ erg$.Given the options,$0.1911 	imes 10^{-10} \ erg$ is the closest value.

In the Bohr model of the atom,when an electron undergoes a transition from $n = 1$ to $n = 3$,the energy absorbed or emitted is ........

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