The ionization enthalpy of a hydrogen atom is | vedclass

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(D) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{1312}{n^2} \, kJ \, mol^{-1} = -\frac{1.312 	imes 10

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) The energy of an electron in the $n^{th}$ orbit of a hydrogen atom is given by $E_n = -\frac{1312}{n^2} \, kJ \, mol^{-1} = -\frac{1.312 	imes 10^6}{n^2} \, J \, mol^{-1}$.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 for excitation is $\Delta E = E_2 - E_1$.$\Delta E = (-0.328 	imes 10^6) - (-1.312 	imes 10^6) = 0.984 	imes 10^6 \, J \, mol^{-1} = 9.84 	imes 10^5 \, J \, mol^{-1}$.

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

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