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प्रश्न
Energy of an electron in the ground state of the hydrogen atom is −2.18 × 10−18 J. Calculate the ionisation enthalpy of atomic hydrogen in terms of kJ mol−1.
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उत्तर
Energy of an electron in the ground state of the hydrogen atom = −2.18 × 10−18 J
\[\ce{H -> H^+ + e^-}\]
The energy required to ionize 1 mole of hydrogen atoms, we multiply by the Avogadro constant.
E = 2.18 × 10−18 × 6.023 × 1023
= 13.123 × 105 J mol−1
I.E = +1312 K J mol−1
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संबंधित प्रश्न
Various successive ionisation enthalpies (in kJ mol-1) of an element are given below.
| IE1 | IE2 | IE3 | IE4 | IE5 |
| 577.5 | 1,810 | 2,750 | 11,580 | 14,820 |
The element is
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