The ionization enthalpy of a hydrogen atom is | vedclass

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(D) The energy of an electron in the $n^{th}$ orbit is given by $E_n = -\frac{1.312 	imes 10^6}{n^2} \ J \ mol^{-1}$.For $n=1$,$E_1 = -1.312 	imes 1

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

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(D) The energy of an electron in the $n^{th}$ orbit is given by $E_n = -\frac{1.312 	imes 10^6}{n^2} \ J \ mol^{-1}$.For $n=1$,$E_1 = -1.312 	imes 10^6 \ J \ mol^{-1}$.For $n=2$,$E_2 = -\frac{1.312 	imes 10^6}{2^2} = -\frac{1.312 	imes 10^6}{4} = -0.328 	imes 10^6 \ J \ mol^{-1}$.The energy required to excite the electron from $n=1$ to $n=2$ is $\Delta E = E_2 - E_1$.$\Delta E = (-0.328 	imes 10^6) - (-1.312 	imes 10^6) \ J \ mol^{-1}$.$\Delta E = (1.312 - 0.328) 	imes 10^6 \ J \ mol^{-1} = 0.984 	imes 10^6 \ J \ mol^{-1} = 9.84 	imes 10^5 \ J \ mol^{-1}$.

The ionization enthalpy of a hydrogen atom is $1.312 \times 10^6 \ J \ mol^{-1}$. The energy required to excite the electron in the atom from $n= 1$ to $n= 2$ is

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