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(B) According to Bohr's 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 energy

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

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(B) According to Bohr's 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 energy change for a transition from $n_1 = 1$ to $n_2 = 3$ is $\Delta E = E_3 - E_1$.$\Delta E = -\frac{2.179 	imes 10^{-11}}{3^2} - (-\frac{2.179 	imes 10^{-11}}{1^2})$.$\Delta E = 2.179 	imes 10^{-11} 	imes (1 - \frac{1}{9}) = 2.179 	imes 10^{-11} 	imes \frac{8}{9}$.$\Delta E = 1.9368 	imes 10^{-11} \ erg = 0.19368 	imes 10^{-10} \ erg$.Comparing with the given options,the closest value is $0.1911 	imes 10^{-10} \ erg$ (Option $B$).Since the electron moves from a lower energy level to a higher energy level,energy is absorbed.

In Bohr's model of an atom,when an electron jumps from $n = 1$ to $n = 3$,how much energy is absorbed?

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