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Energy of an Electron in the Ground State of the Hydrogen Atom is –2.18 × 10–18 J. Calculate the Ionization Enthalpy of Atomic Hydrogen in Terms of J Mol - Chemistry

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

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Solution 1

It is given that the energy of an electron in the ground state of the hydrogen atom is –2.18 × 10–18 J.

Therefore, the energy required to remove that electron from the ground state of hydrogen atom is 2.18 × 10–18 J.

∴ Ionization enthalpy of atomic hydrogen = 2.18 × 10–18 J

Hence, ionization enthalpy of atomic hydrogen in terms of J mol–1 = 2.18 × 10–18 × 6.02 × 1023 J mol–1 = 1.31 × 106 J mol–1

Solution 2

The ionisation enthalpy is for 1 mole atoms. Therefore, ground state energy of the , atoms may be expressed as E (ground state) = ( – 2.18 x 10-18 J) x(6.022 x 1023 mol-1)= -1.312 x 106 J mol-1

Ionisation enthalpy =EE ground state

= 0-(-1.312 x 106mol-1)

= 1.312 x 106 J mol-1.

Concept: Physical Properties - Ionization Enthalpy
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APPEARS IN

NCERT Class 11 Chemistry Textbook
Chapter 3 Classification of Elements and Periodicity in Properties
Q 15 | Page 93
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