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  • Energy of an electron in the ground state of the hydrogen atom is minus 2.18×10 raised to minus 18J. Calculate the ionization enthalpy of atomic hydrogen in terms of J mol raised to minus 1.

3.15   Energy of an electron in the ground state of the hydrogen atom is -2.18 \times 10 ^{-18 }J  Calculate the ionization enthalpy of atomic hydrogen in terms of  J mol ^{-1}

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Given that Energy of an electron in the ground state of the hydrogen atom is -2.18 \times 10 ^{-18 }J.

So, the ionization enthalpy is for 1 mole of atoms.

Therefore, the ground state energy of the atoms may be expressed as :

E (ground state) = -2.18 \times 10 ^{-18 }J \times (6.022\times 10^{23} mol^{-1})

= -1.312\times 10^{6} J\ mol^{-1}

Whereas, ionization enthalpy is, 

=E_{\infty } - E_{ground\ state}

=0-(-1.312\times10^6 mol^{-1}) =1.312\times10^{6} J\ mol^{-1}

Posted by

Divya Prakash Singh

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