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(B) The energy of an electron in the $n$-th orbit of a hydrogen atom is given by $E_n = -\frac{2.18 	imes 10^{-11}}{n^2} \, ergs$. For $n=1$,$E_1 = -

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$0.1911 	imes 10^{-10} \, ergs$

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(B) The energy of an electron in the $n$-th orbit of a hydrogen atom is given by $E_n = -\frac{2.18 	imes 10^{-11}}{n^2} \, ergs$. For $n=1$,$E_1 = -2.18 	imes 10^{-11} \, ergs$. For $n=3$,$E_3 = -\frac{2.18 	imes 10^{-11}}{3^2} = -\frac{2.18 	imes 10^{-11}}{9} \approx -0.2422 	imes 10^{-11} \, ergs$. The energy absorbed is $\Delta E = E_3 - E_1 = (-0.2422 	imes 10^{-11}) - (-2.18 	imes 10^{-11}) = 1.9378 	imes 10^{-11} \, ergs = 0.19378 	imes 10^{-10} \, ergs$. Comparing this with the given options,the closest value is $0.1911 	imes 10^{-10} \, ergs$.

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

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