The energy of an electron in the first Bohr o | vedclass

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(A) The energy of an electron in the $n^{th}$ orbit is given by $E_n = \frac{E_1}{n^2}$,where $E_1 = -13.12 	imes 10^5 \ J \ mol^{-1}$.For the first

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$9.84 	imes 10^5 \ J \ mol^{-1}$

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(A) The energy of an electron in the $n^{th}$ orbit is given by $E_n = \frac{E_1}{n^2}$,where $E_1 = -13.12 	imes 10^5 \ J \ mol^{-1}$.For the first orbit $(n_1 = 1)$,$E_1 = -13.12 	imes 10^5 \ J \ mol^{-1}$.For the second orbit $(n_2 = 2)$,$E_2 = \frac{-13.12 	imes 10^5}{2^2} = \frac{-13.12 	imes 10^5}{4} = -3.28 	imes 10^5 \ J \ mol^{-1}$.The energy required for the transition is $\Delta E = E_2 - E_1$.$\Delta E = (-3.28 	imes 10^5) - (-13.12 	imes 10^5) = 9.84 	imes 10^5 \ J \ mol^{-1}$.

The energy of an electron in the first Bohr orbit of a hydrogen atom is $-13.12 \times 10^5 \ J \ mol^{-1}$. Calculate the energy required for the transition of an electron from the first orbit to the second orbit.

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